Why is the number of ways of choosing 0 items from n items 1?

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It seems easy to grasp that number of ways of choosing $n$ items from $n$ items is 1. But I am unable to understand why is it 1 for choosing 0 items.

combinatorics

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edited 16 mins ago

Glorfindel

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  Is it possible to choose none of the $n$ items? Is there more than one way to do that?
  – bof
  yesterday
  

  
  
  
  


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  The only choice you can make is $lbrace rbrace$
  – P. Quinton
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  What do you think it should be?
  – Martin R
  yesterday
  

  
  
  
  


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  The number of bitstrings of length $n$ containing $n$ ones is the same as the number containing $n$ zeros.
  – bof
  yesterday
  

  
  
  
  


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  How many choices do you have, when choosing zero objects? Only one, take zero of each.
  – AnyAD
  yesterday
  
  

  
  
  
  

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It seems easy to grasp that number of ways of choosing $n$ items from $n$ items is 1. But I am unable to understand why is it 1 for choosing 0 items.

combinatorics

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edited 16 mins ago

Glorfindel

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• 
  
  
  

  
  19
  

  
  
  
  
  
  

  
  Is it possible to choose none of the $n$ items? Is there more than one way to do that?
  – bof
  yesterday
  

  
  
  
  


• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  The only choice you can make is $lbrace rbrace$
  – P. Quinton
  yesterday
  

  
  
  
  


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  4
  

  
  
  
  
  
  

  
  What do you think it should be?
  – Martin R
  yesterday
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  The number of bitstrings of length $n$ containing $n$ ones is the same as the number containing $n$ zeros.
  – bof
  yesterday
  

  
  
  
  


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  1
  

  
  
  
  
  
  

  
  How many choices do you have, when choosing zero objects? Only one, take zero of each.
  – AnyAD
  yesterday
  
  

  
  
  
  

 | 
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It seems easy to grasp that number of ways of choosing $n$ items from $n$ items is 1. But I am unable to understand why is it 1 for choosing 0 items.

combinatorics

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edited 16 mins ago

Glorfindel

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It seems easy to grasp that number of ways of choosing $n$ items from $n$ items is 1. But I am unable to understand why is it 1 for choosing 0 items.

combinatorics

combinatorics

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edited 16 mins ago

Glorfindel

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Glorfindel

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edited 16 mins ago

Glorfindel

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edited 16 mins ago

Glorfindel

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edited 16 mins ago

Glorfindel

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subba

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asked yesterday

subba

20915

asked yesterday

subba

20915

20915

New contributor

subba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor

subba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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• 
  
  
  

  
  19
  

  
  
  
  
  
  

  
  Is it possible to choose none of the $n$ items? Is there more than one way to do that?
  – bof
  yesterday
  

  
  
  
  


• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  The only choice you can make is $lbrace rbrace$
  – P. Quinton
  yesterday
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  What do you think it should be?
  – Martin R
  yesterday
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  The number of bitstrings of length $n$ containing $n$ ones is the same as the number containing $n$ zeros.
  – bof
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  How many choices do you have, when choosing zero objects? Only one, take zero of each.
  – AnyAD
  yesterday
  
  

  
  
  
  

 | 
show 5 more comments

• 
  
  
  

  
  19
  

  
  
  
  
  
  

  
  Is it possible to choose none of the $n$ items? Is there more than one way to do that?
  – bof
  yesterday
  

  
  
  
  


• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  The only choice you can make is $lbrace rbrace$
  – P. Quinton
  yesterday
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  What do you think it should be?
  – Martin R
  yesterday
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  The number of bitstrings of length $n$ containing $n$ ones is the same as the number containing $n$ zeros.
  – bof
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  How many choices do you have, when choosing zero objects? Only one, take zero of each.
  – AnyAD
  yesterday
  
  

  
  
  
  

19

19

Is it possible to choose none of the $n$ items? Is there more than one way to do that?
– bof
yesterday

Is it possible to choose none of the $n$ items? Is there more than one way to do that?
– bof
yesterday

8

8

The only choice you can make is $lbrace rbrace$
– P. Quinton
yesterday

The only choice you can make is $lbrace rbrace$
– P. Quinton
yesterday

4

4

What do you think it should be?
– Martin R
yesterday

What do you think it should be?
– Martin R
yesterday

5

5

The number of bitstrings of length $n$ containing $n$ ones is the same as the number containing $n$ zeros.
– bof
yesterday

The number of bitstrings of length $n$ containing $n$ ones is the same as the number containing $n$ zeros.
– bof
yesterday

1

1

How many choices do you have, when choosing zero objects? Only one, take zero of each.
– AnyAD
yesterday

How many choices do you have, when choosing zero objects? Only one, take zero of each.
– AnyAD
yesterday

 | 
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12 Answers
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When you phrase it in plain English, the answer isn't necessarily totally clear. It might seem reasonable to argue that there are $0$ ways of choosing $0$ things from $n$, since choosing $0$ things isn't really a choice. However, when mathematicians talk about "the number of ways to choose $0$ from $n$," we mean something a bit more specific.

A few ways of describing what we mean:

1. The number of subsets of an $n$-element set with $0$ elements (and we always assume the empty set counts)


2. The number of functions from a $0$-element set to an $n$-element set, up to permutations of the domain (there's exactly one function from the empty set to any set).


3. The coefficient of $x^0y^n$ in the expansion of the binomial $(x+y)^n$.

All of these agree that there is one way to choose $0$ out of $n$ things, and all of these perspectives are mathematically very useful. Thus the perspective that there is a way to choose $0$ out of $n$ things is universal in mathematics.

This wasn't always so obvious to everyone: for instance, I have heard of an early 20th century mathematician who insisted in his books that all intersections be nonempty to be defined. It seems to me that this would be consistent with refusing to accept the empty set as a subset, and with saying there are $0$ ways to choose $0$ things from $n$. But this would make for lots of inconvenient circumlocutions, so we've settled pretty firmly on the other solution.

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answered yesterday

Kevin Carlson

31k23269

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  +1 for the last sentence. There are many examples in math where one could make a reasonable argument for another definition/convention than what is standard, but the standard preserves nice properties that would have to be modified.
  – Carl Schildkraut
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  "It might seem [that] choosing 0 things isn't really a choice." It is if I offer you a plate of cookies and say "Take as many as you want."
  – David Richerby
  6 hours ago
  

  
  
  
  


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  @DavidRicherby An interesting example, because many who make such an offer would not accept choosing to take none!
  – JiK
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  Being an element of the intersection of a collection of sets means being element of every member of the collection. Which would be vacuously true for any candidate element in the case of an empty intersection. So the value of an empty intersection could logically only be a set with everything as element; but we know that such a set cannot exists. Therefore an empty intersection (contrary to an empty union) is not defined; your anonymous mathematician was right.
  – Marc van Leeuwen
  2 hours ago
  

  
  
  
  

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Perhaps a little duality argument can help to provide some intuition.

Given a bag of $n$ balls, consider these two tasks:

1. choose $k$ balls inside the bag, and remove them from the bag


2. choose $n-k$ balls inside the bag, and remove all the others from the bag

Intuition suggests that these tasks are essentially the same: choosing which $k$ balls we remove corresponds to choosing which $n-k$ balls we keep in the bag.

Indeed, there are so many ways to choose $k$ balls (to remove) as there are to choose $n-k$ balls (to keep).

In particular, to determine how many ways we have to choose (and remove) $0$ balls, we can equivalently count how many ways we have to choose (and keep) $n-0=n$ balls. In your own question, you agree on there being only one way to choose $n$ balls out of $n$. Hence, "one way to choose what to keep" can be restated as "one way to choose what to remove".

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answered yesterday

chi

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  I like this intuitive argument. Selecting one from n is equivalent to selecting all but one from n.
  – Nuclear Wang
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  This is especially useful when explaining to a student why nCr = nCn-r.
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Well, since it's possible to choose 0 items from $n$, there must be at least one way to do it. And every way to do it is the same (I admit, this part is harder to formalize), so there is at most one way.

For comparison, there is no way to choose $n+1$ items out of $n$, and equivalently $n choose n+1 = 0$.

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edited yesterday

James Martin

1976

answered yesterday

Glorfindel

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  Alice has a box containing n distinct items from which she is to choose m for inclusion in Bob's (initially empty) box. Carol has a pad of paper, and is instructed to write down a prediction of what Alice will put in Bob's box. She's allowed to write as many predictions as she wants on separate pages. For any given m and n, Dave must construct a table showing the smallest number of pages Carol can use to cover every possible way Alice can set up Bob's box. Dave knows that where _m_=0, Carol can simply turn in a single blank page (or the words "Bob's box is empty".
  – Monty Harder
  21 hours ago
  

  
  
  
  


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  Strictly speaking, I don't think the binomial coefficient is defined for $k > n$.
  – HelloGoodbye
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Your question is equivalent to asking "how many subsets of a set of size of $n$ elements have $0$ elements in it" and we only have $emptyset$.

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answered yesterday

ArsenBerk

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There are several good answers already. I'll try an intuitive explanation. If there are four possible items to buy and several people find none of them attractive enough they will each leave the store with the same contents in their shopping bag. You can't tell the bags apart (assuming they're not personalized ...). There's only one way to buy nothing.

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answered yesterday

Ethan Bolker

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You have written in your question that you understand that "number of ways of choosing n items from n items is 1".
What do you mean by choosing?
Does it mean selecting or does it stands for rejecting ? Or is it just a matter of language and requirement?
Basically it's the latter. And that being said,

number of ways of selecting n items =number of ways of rejecting n items

number of ways of selecting n-1 items =number of ways of rejecting n-1 items

.

.

.

number of ways of selecting 0 items =number of ways of rejecting 0 items

Now,

Logically,

number of ways of rejecting n items =number of ways of selecting 0 items

number of ways of rejecting n-1 items =number of ways of selecting 1 items

.

.

.

number of ways of rejecting 0 items =number of ways of selecting n items

Now ,

number of ways of selecting n items =number of ways of rejecting n items
=1

=>

number of ways of rejecting n items =number of ways of selecting 0 items
=1

Hence Proved!

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answered yesterday

Jalaj Chaturvedi

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  Can you please quote the reason for downvoting? That would really help.
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  This answer is essentially saying the same thing as the answer by @chi and that answer was well received. I up voted this one also. Down-voters, please compare to chi's answer for a different way to look at what this answer is saying.
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You can choose any k items from a bag of n item. And there are different number of ways to choose k items. choosing 0 items from the bag means insert your hand inside the bag and come up with empty hand, just to entertain the kid. You can still do that. So there is no 0 way of doing it. There is still 1 way of doing this.

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answered 10 hours ago

Neel Basu

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Given you have an SO account, I'll add here a general technique for understanding combinatorics that continues to help me daily while doing intense low-level programming.

Bit-permutations can be sub-categorized into bit-combinations to make some counting principles more intuitive. Here's an example of classifying all 256 potential octets (8-bit bytes):

---

0-combinations (8-combinations via dual)

$vertunderline8choose0simeqoverline8choose8vert=underline1+overline1=textbf2$

(00000000) (11111111)

---

1-combinations (7-combinations via dual)

$vertunderline8choose1simeqoverline8choose7vert=underline8+overline8=textbf16$

(00000001) (11111110)
(00000010) (11111101)
(00000100) (11111011)
(00001000) (11110111)
(00010000) (11101111)
(00100000) (11011111)
(01000000) (10111111)
(10000000) (01111111)

---

2-combinations (6-combinations via dual)

$vertunderline8choose2simeqoverline8choose6vert=underline28+overline28=textbf56$

(00000011) (11111100)
(00000101) (11111010)
(00000110) (11111001)
(00001001) (11110110)
(00001010) (11110101)
(00001100) (11110011)
(00010001) (11101110)
(00010010) (11101101)
(00010100) (11101011)
(00011000) (11100111)
(00100001) (11011110)
(00100010) (11011101)
(00100100) (11011011)
(00101000) (11010111)
(00110000) (11001111)
(01000001) (10111110)
(01000010) (10111101)
(01000100) (10111011)
(01001000) (10110111)
(01010000) (10101111)
(01100000) (10011111)
(10000001) (01111110)
(10000010) (01111101)
(10000100) (01111011)
(10001000) (01110111)
(10010000) (01101111)
(10100000) (01011111)
(11000000) (00111111)

---

3-combinations (5-combinations via dual)

$vertunderline8choose3simeqoverline8choose5vert=underline56+overline56=textbf112$

(00000111) (11111000)
(00001011) (11110100)
(00001101) (11110010)
(00001110) (11110001)
(00010011) (11101100)
(00010101) (11101010)
(00010110) (11101001)
(00011001) (11100110)
(00011010) (11100101)
(00011100) (11100011)
(00100011) (11011100)
(00100101) (11011010)
(00100110) (11011001)
(00101001) (11010110)
(00101010) (11010101)
(00101100) (11010011)
(00110001) (11001110)
(00110010) (11001101)
(00110100) (11001011)
(00111000) (11000111)
(01000011) (10111100)
(01000101) (10111010)
(01000110) (10111001)
(01001001) (10110110)
(01001010) (10110101)
(01001100) (10110011)
(01010001) (10101110)
(01010010) (10101101)
(01010100) (10101011)
(01011000) (10100111)
(01100001) (10011110)
(01100010) (10011101)
(01100100) (10011011)
(01101000) (10010111)
(01110000) (10001111)
(10000011) (01111100)
(10000101) (01111010)
(10000110) (01111001)
(10001001) (01110110)
(10001010) (01110101)
(10001100) (01110011)
(10010001) (01101110)
(10010010) (01101101)
(10010100) (01101011)
(10011000) (01100111)
(10100001) (01011110)
(10100010) (01011101)
(10100100) (01011011)
(10101000) (01010111)
(10110000) (01001111)
(11000001) (00111110)
(11000010) (00111101)
(11000100) (00111011)
(11001000) (00110111)
(11010000) (00101111)
(11100000) (00011111)

---

4-combinations (self-dual)

$vertunderlineoverline8choose4vert=underlineoverline70=textbf70$

(00001111)
(00010111)
(00011011)
(00011101)
(00011110)
(00100111)
(00101011)
(00101101)
(00101110)
(00110011)
(00110101)
(00110110)
(00111001)
(00111010)
(00111100)
(01000111)
(01001011)
(01001101)
(01001110)
(01010011)
(01010101)
(01010110)
(01011001)
(01011010)
(01011100)
(01100011)
(01100101)
(01100110)
(01101001)
(01101010)
(01101100)
(01110001)
(01110010)
(01110100)
(01111000)
(10000111)
(10001011)
(10001101)
(10001110)
(10010011)
(10010101)
(10010110)
(10011001)
(10011010)
(10011100)
(10100011)
(10100101)
(10100110)
(10101001)
(10101010)
(10101100)
(10110001)
(10110010)
(10110100)
(10111000)
(11000011)
(11000101)
(11000110)
(11001001)
(11001010)
(11001100)
(11010001)
(11010010)
(11010100)
(11011000)
(11100001)
(11100010)
(11100100)
(11101000)
(11110000)

---

$vertPermutationsvert = 2+16+56+112+70=textbf256$

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answered 12 hours ago

user13972

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up vote
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down vote

Imaginate as an option:

1) Pick $0$ elements

2) Pick $1$ element

3) Pick $2$ elements

...

n+1) Pick $n$ elements

---

(EDIT)
Life examples:

I. Voting: You can choose between Candidate A, Candidate B or no one by not voting.

Note that choosing nobody is a choose as well (not on paper of course).

II. Shopping: You want to buy a specific type of bread (like: gluten free)- so if there is- you pick or if there is no- 'that's your default vallue'.

Your set here is $text0,1$ with the cardinality of $2$ results.

III. A school mate in 2005:

"As you see (Chris) I have ten fingers: 1,2,3...10.

Let's count now from zero to be sure: 0,1,2...9"

This puzzle is not the best sample (since he could start to counter by minus one for example) but the point was: Finger #$0$ is a finger as well .

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edited 12 hours ago

answered yesterday

Krzysztof Myśliwiec

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  Could you elaborate, please?
  – DaG
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  @DaG Yes- tommorrow I made an edit ( hopefully without english grammar mistakes).
  – Krzysztof Myśliwiec
  yesterday
  

  
  
  
  


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  @DaG I not unclude terms like the Empty Set to not complicate my primary answer
  – Krzysztof Myśliwiec
  12 hours ago
  

  
  
  
  

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up vote
1
down vote

Think of it like this: Choose the items that you don't want to choose. By choosing $k$ items out of $n$ not to choose, you will effectively have chosen $n - k$ out of the $n$ items, and it becomes obvious that there are equally many ways of choosing $k$ items out of $n$ as there are ways of choosing $n-k$ items out of $n$.

So, there are equally many ways of choosing 0 items out of $n$ as there are ways of choosing $n$ items out of $n$.

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answered 10 hours ago

HelloGoodbye

1905

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up vote
1
down vote

The other answers are fine, but for a geometric twist:

Pascal's triangle would be a lot less elegant without this definition:

 1
 1 1
 1 2 1
 1 3 3 1
1 4 6 4 1

(And so on.) Even if there wasn't any other compelling reason for defining $n choose k$ in such a way that $nchoose 0 = 0$, the symmetry of the triangle itself could lead you there. It would be odd if the last 4 entries of the bottom row were $4choose 1, ldots, 4 choose 4$ but the first entry wasn't $4 choose 0$.

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answered 3 hours ago

John Coleman

3,66811123

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up vote
1
down vote

How to choose 0 items: Choose n items, then pick the ones that remain. There is one way to choose n items, therefore there is only one way to pick the ones that remain.

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answered 2 hours ago

gnasher729

5,9111028

add a comment | 

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up vote
59
down vote



accepted

When you phrase it in plain English, the answer isn't necessarily totally clear. It might seem reasonable to argue that there are $0$ ways of choosing $0$ things from $n$, since choosing $0$ things isn't really a choice. However, when mathematicians talk about "the number of ways to choose $0$ from $n$," we mean something a bit more specific.

A few ways of describing what we mean:

1. The number of subsets of an $n$-element set with $0$ elements (and we always assume the empty set counts)


2. The number of functions from a $0$-element set to an $n$-element set, up to permutations of the domain (there's exactly one function from the empty set to any set).


3. The coefficient of $x^0y^n$ in the expansion of the binomial $(x+y)^n$.

All of these agree that there is one way to choose $0$ out of $n$ things, and all of these perspectives are mathematically very useful. Thus the perspective that there is a way to choose $0$ out of $n$ things is universal in mathematics.

This wasn't always so obvious to everyone: for instance, I have heard of an early 20th century mathematician who insisted in his books that all intersections be nonempty to be defined. It seems to me that this would be consistent with refusing to accept the empty set as a subset, and with saying there are $0$ ways to choose $0$ things from $n$. But this would make for lots of inconvenient circumlocutions, so we've settled pretty firmly on the other solution.

share|cite|improve this answer

answered yesterday

Kevin Carlson

31k23269

• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  +1 for the last sentence. There are many examples in math where one could make a reasonable argument for another definition/convention than what is standard, but the standard preserves nice properties that would have to be modified.
  – Carl Schildkraut
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  "It might seem [that] choosing 0 things isn't really a choice." It is if I offer you a plate of cookies and say "Take as many as you want."
  – David Richerby
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidRicherby An interesting example, because many who make such an offer would not accept choosing to take none!
  – JiK
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Being an element of the intersection of a collection of sets means being element of every member of the collection. Which would be vacuously true for any candidate element in the case of an empty intersection. So the value of an empty intersection could logically only be a set with everything as element; but we know that such a set cannot exists. Therefore an empty intersection (contrary to an empty union) is not defined; your anonymous mathematician was right.
  – Marc van Leeuwen
  2 hours ago
  

  
  
  
  

add a comment | 

up vote
59
down vote



accepted

When you phrase it in plain English, the answer isn't necessarily totally clear. It might seem reasonable to argue that there are $0$ ways of choosing $0$ things from $n$, since choosing $0$ things isn't really a choice. However, when mathematicians talk about "the number of ways to choose $0$ from $n$," we mean something a bit more specific.

A few ways of describing what we mean:

1. The number of subsets of an $n$-element set with $0$ elements (and we always assume the empty set counts)


2. The number of functions from a $0$-element set to an $n$-element set, up to permutations of the domain (there's exactly one function from the empty set to any set).


3. The coefficient of $x^0y^n$ in the expansion of the binomial $(x+y)^n$.

All of these agree that there is one way to choose $0$ out of $n$ things, and all of these perspectives are mathematically very useful. Thus the perspective that there is a way to choose $0$ out of $n$ things is universal in mathematics.

This wasn't always so obvious to everyone: for instance, I have heard of an early 20th century mathematician who insisted in his books that all intersections be nonempty to be defined. It seems to me that this would be consistent with refusing to accept the empty set as a subset, and with saying there are $0$ ways to choose $0$ things from $n$. But this would make for lots of inconvenient circumlocutions, so we've settled pretty firmly on the other solution.

share|cite|improve this answer

answered yesterday

Kevin Carlson

31k23269

• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  +1 for the last sentence. There are many examples in math where one could make a reasonable argument for another definition/convention than what is standard, but the standard preserves nice properties that would have to be modified.
  – Carl Schildkraut
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  "It might seem [that] choosing 0 things isn't really a choice." It is if I offer you a plate of cookies and say "Take as many as you want."
  – David Richerby
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidRicherby An interesting example, because many who make such an offer would not accept choosing to take none!
  – JiK
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Being an element of the intersection of a collection of sets means being element of every member of the collection. Which would be vacuously true for any candidate element in the case of an empty intersection. So the value of an empty intersection could logically only be a set with everything as element; but we know that such a set cannot exists. Therefore an empty intersection (contrary to an empty union) is not defined; your anonymous mathematician was right.
  – Marc van Leeuwen
  2 hours ago
  

  
  
  
  

add a comment | 

up vote
59
down vote



accepted

up vote
59
down vote



accepted

When you phrase it in plain English, the answer isn't necessarily totally clear. It might seem reasonable to argue that there are $0$ ways of choosing $0$ things from $n$, since choosing $0$ things isn't really a choice. However, when mathematicians talk about "the number of ways to choose $0$ from $n$," we mean something a bit more specific.

A few ways of describing what we mean:

1. The number of subsets of an $n$-element set with $0$ elements (and we always assume the empty set counts)


2. The number of functions from a $0$-element set to an $n$-element set, up to permutations of the domain (there's exactly one function from the empty set to any set).


3. The coefficient of $x^0y^n$ in the expansion of the binomial $(x+y)^n$.

All of these agree that there is one way to choose $0$ out of $n$ things, and all of these perspectives are mathematically very useful. Thus the perspective that there is a way to choose $0$ out of $n$ things is universal in mathematics.

This wasn't always so obvious to everyone: for instance, I have heard of an early 20th century mathematician who insisted in his books that all intersections be nonempty to be defined. It seems to me that this would be consistent with refusing to accept the empty set as a subset, and with saying there are $0$ ways to choose $0$ things from $n$. But this would make for lots of inconvenient circumlocutions, so we've settled pretty firmly on the other solution.

share|cite|improve this answer

answered yesterday

Kevin Carlson

31k23269

When you phrase it in plain English, the answer isn't necessarily totally clear. It might seem reasonable to argue that there are $0$ ways of choosing $0$ things from $n$, since choosing $0$ things isn't really a choice. However, when mathematicians talk about "the number of ways to choose $0$ from $n$," we mean something a bit more specific.

A few ways of describing what we mean:

1. The number of subsets of an $n$-element set with $0$ elements (and we always assume the empty set counts)


2. The number of functions from a $0$-element set to an $n$-element set, up to permutations of the domain (there's exactly one function from the empty set to any set).


3. The coefficient of $x^0y^n$ in the expansion of the binomial $(x+y)^n$.

All of these agree that there is one way to choose $0$ out of $n$ things, and all of these perspectives are mathematically very useful. Thus the perspective that there is a way to choose $0$ out of $n$ things is universal in mathematics.

This wasn't always so obvious to everyone: for instance, I have heard of an early 20th century mathematician who insisted in his books that all intersections be nonempty to be defined. It seems to me that this would be consistent with refusing to accept the empty set as a subset, and with saying there are $0$ ways to choose $0$ things from $n$. But this would make for lots of inconvenient circumlocutions, so we've settled pretty firmly on the other solution.

share|cite|improve this answer

answered yesterday

Kevin Carlson

31k23269

share|cite|improve this answer

share|cite|improve this answer

answered yesterday

Kevin Carlson

31k23269

answered yesterday

Kevin Carlson

31k23269

answered yesterday

Kevin Carlson

31k23269

31k23269

• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  +1 for the last sentence. There are many examples in math where one could make a reasonable argument for another definition/convention than what is standard, but the standard preserves nice properties that would have to be modified.
  – Carl Schildkraut
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  "It might seem [that] choosing 0 things isn't really a choice." It is if I offer you a plate of cookies and say "Take as many as you want."
  – David Richerby
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidRicherby An interesting example, because many who make such an offer would not accept choosing to take none!
  – JiK
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Being an element of the intersection of a collection of sets means being element of every member of the collection. Which would be vacuously true for any candidate element in the case of an empty intersection. So the value of an empty intersection could logically only be a set with everything as element; but we know that such a set cannot exists. Therefore an empty intersection (contrary to an empty union) is not defined; your anonymous mathematician was right.
  – Marc van Leeuwen
  2 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  +1 for the last sentence. There are many examples in math where one could make a reasonable argument for another definition/convention than what is standard, but the standard preserves nice properties that would have to be modified.
  – Carl Schildkraut
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  "It might seem [that] choosing 0 things isn't really a choice." It is if I offer you a plate of cookies and say "Take as many as you want."
  – David Richerby
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidRicherby An interesting example, because many who make such an offer would not accept choosing to take none!
  – JiK
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Being an element of the intersection of a collection of sets means being element of every member of the collection. Which would be vacuously true for any candidate element in the case of an empty intersection. So the value of an empty intersection could logically only be a set with everything as element; but we know that such a set cannot exists. Therefore an empty intersection (contrary to an empty union) is not defined; your anonymous mathematician was right.
  – Marc van Leeuwen
  2 hours ago
  

  
  
  
  

7

7

+1 for the last sentence. There are many examples in math where one could make a reasonable argument for another definition/convention than what is standard, but the standard preserves nice properties that would have to be modified.
– Carl Schildkraut
21 hours ago

+1 for the last sentence. There are many examples in math where one could make a reasonable argument for another definition/convention than what is standard, but the standard preserves nice properties that would have to be modified.
– Carl Schildkraut
21 hours ago

4

4

"It might seem [that] choosing 0 things isn't really a choice." It is if I offer you a plate of cookies and say "Take as many as you want."
– David Richerby
6 hours ago

"It might seem [that] choosing 0 things isn't really a choice." It is if I offer you a plate of cookies and say "Take as many as you want."
– David Richerby
6 hours ago

@DavidRicherby An interesting example, because many who make such an offer would not accept choosing to take none!
– JiK
3 hours ago

@DavidRicherby An interesting example, because many who make such an offer would not accept choosing to take none!
– JiK
3 hours ago

Being an element of the intersection of a collection of sets means being element of every member of the collection. Which would be vacuously true for any candidate element in the case of an empty intersection. So the value of an empty intersection could logically only be a set with everything as element; but we know that such a set cannot exists. Therefore an empty intersection (contrary to an empty union) is not defined; your anonymous mathematician was right.
– Marc van Leeuwen
2 hours ago

Being an element of the intersection of a collection of sets means being element of every member of the collection. Which would be vacuously true for any candidate element in the case of an empty intersection. So the value of an empty intersection could logically only be a set with everything as element; but we know that such a set cannot exists. Therefore an empty intersection (contrary to an empty union) is not defined; your anonymous mathematician was right.
– Marc van Leeuwen
2 hours ago

add a comment | 

up vote
38
down vote

Perhaps a little duality argument can help to provide some intuition.

Given a bag of $n$ balls, consider these two tasks:

1. choose $k$ balls inside the bag, and remove them from the bag


2. choose $n-k$ balls inside the bag, and remove all the others from the bag

Intuition suggests that these tasks are essentially the same: choosing which $k$ balls we remove corresponds to choosing which $n-k$ balls we keep in the bag.

Indeed, there are so many ways to choose $k$ balls (to remove) as there are to choose $n-k$ balls (to keep).

In particular, to determine how many ways we have to choose (and remove) $0$ balls, we can equivalently count how many ways we have to choose (and keep) $n-0=n$ balls. In your own question, you agree on there being only one way to choose $n$ balls out of $n$. Hence, "one way to choose what to keep" can be restated as "one way to choose what to remove".

share|cite|improve this answer

answered yesterday

chi

1,284713

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  I like this intuitive argument. Selecting one from n is equivalent to selecting all but one from n.
  – Nuclear Wang
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This is especially useful when explaining to a student why nCr = nCn-r.
  – nurdyguy
  41 mins ago
  

  
  
  
  

add a comment | 

up vote
38
down vote

Perhaps a little duality argument can help to provide some intuition.

Given a bag of $n$ balls, consider these two tasks:

1. choose $k$ balls inside the bag, and remove them from the bag


2. choose $n-k$ balls inside the bag, and remove all the others from the bag

Intuition suggests that these tasks are essentially the same: choosing which $k$ balls we remove corresponds to choosing which $n-k$ balls we keep in the bag.

Indeed, there are so many ways to choose $k$ balls (to remove) as there are to choose $n-k$ balls (to keep).

In particular, to determine how many ways we have to choose (and remove) $0$ balls, we can equivalently count how many ways we have to choose (and keep) $n-0=n$ balls. In your own question, you agree on there being only one way to choose $n$ balls out of $n$. Hence, "one way to choose what to keep" can be restated as "one way to choose what to remove".

share|cite|improve this answer

answered yesterday

chi

1,284713

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  I like this intuitive argument. Selecting one from n is equivalent to selecting all but one from n.
  – Nuclear Wang
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This is especially useful when explaining to a student why nCr = nCn-r.
  – nurdyguy
  41 mins ago
  

  
  
  
  

add a comment | 

up vote
38
down vote

up vote
38
down vote

Perhaps a little duality argument can help to provide some intuition.

Given a bag of $n$ balls, consider these two tasks:

1. choose $k$ balls inside the bag, and remove them from the bag


2. choose $n-k$ balls inside the bag, and remove all the others from the bag

Intuition suggests that these tasks are essentially the same: choosing which $k$ balls we remove corresponds to choosing which $n-k$ balls we keep in the bag.

Indeed, there are so many ways to choose $k$ balls (to remove) as there are to choose $n-k$ balls (to keep).

In particular, to determine how many ways we have to choose (and remove) $0$ balls, we can equivalently count how many ways we have to choose (and keep) $n-0=n$ balls. In your own question, you agree on there being only one way to choose $n$ balls out of $n$. Hence, "one way to choose what to keep" can be restated as "one way to choose what to remove".

share|cite|improve this answer

answered yesterday

chi

1,284713

Perhaps a little duality argument can help to provide some intuition.

Given a bag of $n$ balls, consider these two tasks:

1. choose $k$ balls inside the bag, and remove them from the bag


2. choose $n-k$ balls inside the bag, and remove all the others from the bag

Intuition suggests that these tasks are essentially the same: choosing which $k$ balls we remove corresponds to choosing which $n-k$ balls we keep in the bag.

Indeed, there are so many ways to choose $k$ balls (to remove) as there are to choose $n-k$ balls (to keep).

In particular, to determine how many ways we have to choose (and remove) $0$ balls, we can equivalently count how many ways we have to choose (and keep) $n-0=n$ balls. In your own question, you agree on there being only one way to choose $n$ balls out of $n$. Hence, "one way to choose what to keep" can be restated as "one way to choose what to remove".

share|cite|improve this answer

answered yesterday

chi

1,284713

share|cite|improve this answer

share|cite|improve this answer

answered yesterday

chi

1,284713

answered yesterday

chi

1,284713

answered yesterday

chi

1,284713

1,284713

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  I like this intuitive argument. Selecting one from n is equivalent to selecting all but one from n.
  – Nuclear Wang
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This is especially useful when explaining to a student why nCr = nCn-r.
  – nurdyguy
  41 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  I like this intuitive argument. Selecting one from n is equivalent to selecting all but one from n.
  – Nuclear Wang
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This is especially useful when explaining to a student why nCr = nCn-r.
  – nurdyguy
  41 mins ago
  

  
  
  
  

2

2

I like this intuitive argument. Selecting one from n is equivalent to selecting all but one from n.
– Nuclear Wang
yesterday

I like this intuitive argument. Selecting one from n is equivalent to selecting all but one from n.
– Nuclear Wang
yesterday

1

1

This is especially useful when explaining to a student why nCr = nCn-r.
– nurdyguy
41 mins ago

This is especially useful when explaining to a student why nCr = nCn-r.
– nurdyguy
41 mins ago

add a comment | 

up vote
9
down vote

Well, since it's possible to choose 0 items from $n$, there must be at least one way to do it. And every way to do it is the same (I admit, this part is harder to formalize), so there is at most one way.

For comparison, there is no way to choose $n+1$ items out of $n$, and equivalently $n choose n+1 = 0$.

share|cite|improve this answer

edited yesterday

James Martin

1976

answered yesterday

Glorfindel

3,33471729

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Alice has a box containing n distinct items from which she is to choose m for inclusion in Bob's (initially empty) box. Carol has a pad of paper, and is instructed to write down a prediction of what Alice will put in Bob's box. She's allowed to write as many predictions as she wants on separate pages. For any given m and n, Dave must construct a table showing the smallest number of pages Carol can use to cover every possible way Alice can set up Bob's box. Dave knows that where _m_=0, Carol can simply turn in a single blank page (or the words "Bob's box is empty".
  – Monty Harder
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Strictly speaking, I don't think the binomial coefficient is defined for $k > n$.
  – HelloGoodbye
  10 hours ago
  

  
  
  
  

add a comment | 

up vote
9
down vote

Well, since it's possible to choose 0 items from $n$, there must be at least one way to do it. And every way to do it is the same (I admit, this part is harder to formalize), so there is at most one way.

For comparison, there is no way to choose $n+1$ items out of $n$, and equivalently $n choose n+1 = 0$.

share|cite|improve this answer

edited yesterday

James Martin

1976

answered yesterday

Glorfindel

3,33471729

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Alice has a box containing n distinct items from which she is to choose m for inclusion in Bob's (initially empty) box. Carol has a pad of paper, and is instructed to write down a prediction of what Alice will put in Bob's box. She's allowed to write as many predictions as she wants on separate pages. For any given m and n, Dave must construct a table showing the smallest number of pages Carol can use to cover every possible way Alice can set up Bob's box. Dave knows that where _m_=0, Carol can simply turn in a single blank page (or the words "Bob's box is empty".
  – Monty Harder
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Strictly speaking, I don't think the binomial coefficient is defined for $k > n$.
  – HelloGoodbye
  10 hours ago
  

  
  
  
  

add a comment | 

up vote
9
down vote

up vote
9
down vote

Well, since it's possible to choose 0 items from $n$, there must be at least one way to do it. And every way to do it is the same (I admit, this part is harder to formalize), so there is at most one way.

For comparison, there is no way to choose $n+1$ items out of $n$, and equivalently $n choose n+1 = 0$.

share|cite|improve this answer

edited yesterday

James Martin

1976

answered yesterday

Glorfindel

3,33471729

Well, since it's possible to choose 0 items from $n$, there must be at least one way to do it. And every way to do it is the same (I admit, this part is harder to formalize), so there is at most one way.

For comparison, there is no way to choose $n+1$ items out of $n$, and equivalently $n choose n+1 = 0$.

share|cite|improve this answer

edited yesterday

James Martin

1976

answered yesterday

Glorfindel

3,33471729

share|cite|improve this answer

share|cite|improve this answer

edited yesterday

James Martin

1976

edited yesterday

James Martin

1976

edited yesterday

James Martin

1976

1976

answered yesterday

Glorfindel

3,33471729

answered yesterday

Glorfindel

3,33471729

answered yesterday

Glorfindel

3,33471729

3,33471729

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Alice has a box containing n distinct items from which she is to choose m for inclusion in Bob's (initially empty) box. Carol has a pad of paper, and is instructed to write down a prediction of what Alice will put in Bob's box. She's allowed to write as many predictions as she wants on separate pages. For any given m and n, Dave must construct a table showing the smallest number of pages Carol can use to cover every possible way Alice can set up Bob's box. Dave knows that where _m_=0, Carol can simply turn in a single blank page (or the words "Bob's box is empty".
  – Monty Harder
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Strictly speaking, I don't think the binomial coefficient is defined for $k > n$.
  – HelloGoodbye
  10 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Alice has a box containing n distinct items from which she is to choose m for inclusion in Bob's (initially empty) box. Carol has a pad of paper, and is instructed to write down a prediction of what Alice will put in Bob's box. She's allowed to write as many predictions as she wants on separate pages. For any given m and n, Dave must construct a table showing the smallest number of pages Carol can use to cover every possible way Alice can set up Bob's box. Dave knows that where _m_=0, Carol can simply turn in a single blank page (or the words "Bob's box is empty".
  – Monty Harder
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Strictly speaking, I don't think the binomial coefficient is defined for $k > n$.
  – HelloGoodbye
  10 hours ago
  

  
  
  
  

Alice has a box containing n distinct items from which she is to choose m for inclusion in Bob's (initially empty) box. Carol has a pad of paper, and is instructed to write down a prediction of what Alice will put in Bob's box. She's allowed to write as many predictions as she wants on separate pages. For any given m and n, Dave must construct a table showing the smallest number of pages Carol can use to cover every possible way Alice can set up Bob's box. Dave knows that where _m_=0, Carol can simply turn in a single blank page (or the words "Bob's box is empty".
– Monty Harder
21 hours ago

Alice has a box containing n distinct items from which she is to choose m for inclusion in Bob's (initially empty) box. Carol has a pad of paper, and is instructed to write down a prediction of what Alice will put in Bob's box. She's allowed to write as many predictions as she wants on separate pages. For any given m and n, Dave must construct a table showing the smallest number of pages Carol can use to cover every possible way Alice can set up Bob's box. Dave knows that where _m_=0, Carol can simply turn in a single blank page (or the words "Bob's box is empty".
– Monty Harder
21 hours ago

Strictly speaking, I don't think the binomial coefficient is defined for $k > n$.
– HelloGoodbye
10 hours ago

Strictly speaking, I don't think the binomial coefficient is defined for $k > n$.
– HelloGoodbye
10 hours ago

add a comment | 

up vote
8
down vote

Your question is equivalent to asking "how many subsets of a set of size of $n$ elements have $0$ elements in it" and we only have $emptyset$.

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answered yesterday

ArsenBerk

7,47121234

add a comment | 

up vote
8
down vote

Your question is equivalent to asking "how many subsets of a set of size of $n$ elements have $0$ elements in it" and we only have $emptyset$.

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answered yesterday

ArsenBerk

7,47121234

add a comment | 

up vote
8
down vote

up vote
8
down vote

Your question is equivalent to asking "how many subsets of a set of size of $n$ elements have $0$ elements in it" and we only have $emptyset$.

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answered yesterday

ArsenBerk

7,47121234

Your question is equivalent to asking "how many subsets of a set of size of $n$ elements have $0$ elements in it" and we only have $emptyset$.

share|cite|improve this answer

answered yesterday

ArsenBerk

7,47121234

share|cite|improve this answer

share|cite|improve this answer

answered yesterday

ArsenBerk

7,47121234

answered yesterday

ArsenBerk

7,47121234

answered yesterday

ArsenBerk

7,47121234

7,47121234

add a comment | 

add a comment | 

up vote
5
down vote

There are several good answers already. I'll try an intuitive explanation. If there are four possible items to buy and several people find none of them attractive enough they will each leave the store with the same contents in their shopping bag. You can't tell the bags apart (assuming they're not personalized ...). There's only one way to buy nothing.

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answered yesterday

Ethan Bolker

38k543100

add a comment | 

up vote
5
down vote

There are several good answers already. I'll try an intuitive explanation. If there are four possible items to buy and several people find none of them attractive enough they will each leave the store with the same contents in their shopping bag. You can't tell the bags apart (assuming they're not personalized ...). There's only one way to buy nothing.

share|cite|improve this answer

answered yesterday

Ethan Bolker

38k543100

add a comment | 

up vote
5
down vote

up vote
5
down vote

There are several good answers already. I'll try an intuitive explanation. If there are four possible items to buy and several people find none of them attractive enough they will each leave the store with the same contents in their shopping bag. You can't tell the bags apart (assuming they're not personalized ...). There's only one way to buy nothing.

share|cite|improve this answer

answered yesterday

Ethan Bolker

38k543100

There are several good answers already. I'll try an intuitive explanation. If there are four possible items to buy and several people find none of them attractive enough they will each leave the store with the same contents in their shopping bag. You can't tell the bags apart (assuming they're not personalized ...). There's only one way to buy nothing.

share|cite|improve this answer

answered yesterday

Ethan Bolker

38k543100

share|cite|improve this answer

share|cite|improve this answer

answered yesterday

Ethan Bolker

38k543100

answered yesterday

Ethan Bolker

38k543100

answered yesterday

Ethan Bolker

38k543100

38k543100

add a comment | 

add a comment | 

up vote
3
down vote

You have written in your question that you understand that "number of ways of choosing n items from n items is 1".
What do you mean by choosing?
Does it mean selecting or does it stands for rejecting ? Or is it just a matter of language and requirement?
Basically it's the latter. And that being said,

number of ways of selecting n items =number of ways of rejecting n items

number of ways of selecting n-1 items =number of ways of rejecting n-1 items

.

.

.

number of ways of selecting 0 items =number of ways of rejecting 0 items

Now,

Logically,

number of ways of rejecting n items =number of ways of selecting 0 items

number of ways of rejecting n-1 items =number of ways of selecting 1 items

.

.

.

number of ways of rejecting 0 items =number of ways of selecting n items

Now ,

number of ways of selecting n items =number of ways of rejecting n items
=1

=>

number of ways of rejecting n items =number of ways of selecting 0 items
=1

Hence Proved!

share|cite|improve this answer

answered yesterday

Jalaj Chaturvedi

411

New contributor

Jalaj Chaturvedi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you please quote the reason for downvoting? That would really help.
  – Jalaj Chaturvedi
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This answer is essentially saying the same thing as the answer by @chi and that answer was well received. I up voted this one also. Down-voters, please compare to chi's answer for a different way to look at what this answer is saying.
  – Aaron
  yesterday
  

  
  
  
  

add a comment | 

up vote
3
down vote

You have written in your question that you understand that "number of ways of choosing n items from n items is 1".
What do you mean by choosing?
Does it mean selecting or does it stands for rejecting ? Or is it just a matter of language and requirement?
Basically it's the latter. And that being said,

number of ways of selecting n items =number of ways of rejecting n items

number of ways of selecting n-1 items =number of ways of rejecting n-1 items

.

.

.

number of ways of selecting 0 items =number of ways of rejecting 0 items

Now,

Logically,

number of ways of rejecting n items =number of ways of selecting 0 items

number of ways of rejecting n-1 items =number of ways of selecting 1 items

.

.

.

number of ways of rejecting 0 items =number of ways of selecting n items

Now ,

number of ways of selecting n items =number of ways of rejecting n items
=1

=>

number of ways of rejecting n items =number of ways of selecting 0 items
=1

Hence Proved!

share|cite|improve this answer

answered yesterday

Jalaj Chaturvedi

411

New contributor

Jalaj Chaturvedi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you please quote the reason for downvoting? That would really help.
  – Jalaj Chaturvedi
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This answer is essentially saying the same thing as the answer by @chi and that answer was well received. I up voted this one also. Down-voters, please compare to chi's answer for a different way to look at what this answer is saying.
  – Aaron
  yesterday
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

You have written in your question that you understand that "number of ways of choosing n items from n items is 1".
What do you mean by choosing?
Does it mean selecting or does it stands for rejecting ? Or is it just a matter of language and requirement?
Basically it's the latter. And that being said,

number of ways of selecting n items =number of ways of rejecting n items

number of ways of selecting n-1 items =number of ways of rejecting n-1 items

.

.

.

number of ways of selecting 0 items =number of ways of rejecting 0 items

Now,

Logically,

number of ways of rejecting n items =number of ways of selecting 0 items

number of ways of rejecting n-1 items =number of ways of selecting 1 items

.

.

.

number of ways of rejecting 0 items =number of ways of selecting n items

Now ,

number of ways of selecting n items =number of ways of rejecting n items
=1

=>

number of ways of rejecting n items =number of ways of selecting 0 items
=1

Hence Proved!

share|cite|improve this answer

answered yesterday

Jalaj Chaturvedi

411

New contributor

Jalaj Chaturvedi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

You have written in your question that you understand that "number of ways of choosing n items from n items is 1".
What do you mean by choosing?
Does it mean selecting or does it stands for rejecting ? Or is it just a matter of language and requirement?
Basically it's the latter. And that being said,

number of ways of selecting n items =number of ways of rejecting n items

number of ways of selecting n-1 items =number of ways of rejecting n-1 items

.

.

.

number of ways of selecting 0 items =number of ways of rejecting 0 items

Now,

Logically,

number of ways of rejecting n items =number of ways of selecting 0 items

number of ways of rejecting n-1 items =number of ways of selecting 1 items

.

.

.

number of ways of rejecting 0 items =number of ways of selecting n items

Now ,

number of ways of selecting n items =number of ways of rejecting n items
=1

=>

number of ways of rejecting n items =number of ways of selecting 0 items
=1

Hence Proved!

share|cite|improve this answer

answered yesterday

Jalaj Chaturvedi

411

New contributor

Jalaj Chaturvedi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this answer

share|cite|improve this answer

answered yesterday

Jalaj Chaturvedi

411

New contributor

Jalaj Chaturvedi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered yesterday

Jalaj Chaturvedi

411

answered yesterday

Jalaj Chaturvedi

411

411

New contributor

Jalaj Chaturvedi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

Jalaj Chaturvedi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Jalaj Chaturvedi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you please quote the reason for downvoting? That would really help.
  – Jalaj Chaturvedi
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This answer is essentially saying the same thing as the answer by @chi and that answer was well received. I up voted this one also. Down-voters, please compare to chi's answer for a different way to look at what this answer is saying.
  – Aaron
  yesterday
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you please quote the reason for downvoting? That would really help.
  – Jalaj Chaturvedi
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This answer is essentially saying the same thing as the answer by @chi and that answer was well received. I up voted this one also. Down-voters, please compare to chi's answer for a different way to look at what this answer is saying.
  – Aaron
  yesterday
  

  
  
  
  

Can you please quote the reason for downvoting? That would really help.
– Jalaj Chaturvedi
yesterday

Can you please quote the reason for downvoting? That would really help.
– Jalaj Chaturvedi
yesterday

This answer is essentially saying the same thing as the answer by @chi and that answer was well received. I up voted this one also. Down-voters, please compare to chi's answer for a different way to look at what this answer is saying.
– Aaron
yesterday

This answer is essentially saying the same thing as the answer by @chi and that answer was well received. I up voted this one also. Down-voters, please compare to chi's answer for a different way to look at what this answer is saying.
– Aaron
yesterday

add a comment | 

up vote
2
down vote

You can choose any k items from a bag of n item. And there are different number of ways to choose k items. choosing 0 items from the bag means insert your hand inside the bag and come up with empty hand, just to entertain the kid. You can still do that. So there is no 0 way of doing it. There is still 1 way of doing this.

share|cite|improve this answer

answered 10 hours ago

Neel Basu

19510

add a comment | 

up vote
2
down vote

You can choose any k items from a bag of n item. And there are different number of ways to choose k items. choosing 0 items from the bag means insert your hand inside the bag and come up with empty hand, just to entertain the kid. You can still do that. So there is no 0 way of doing it. There is still 1 way of doing this.

share|cite|improve this answer

answered 10 hours ago

Neel Basu

19510

add a comment | 

up vote
2
down vote

up vote
2
down vote

You can choose any k items from a bag of n item. And there are different number of ways to choose k items. choosing 0 items from the bag means insert your hand inside the bag and come up with empty hand, just to entertain the kid. You can still do that. So there is no 0 way of doing it. There is still 1 way of doing this.

share|cite|improve this answer

answered 10 hours ago

Neel Basu

19510

You can choose any k items from a bag of n item. And there are different number of ways to choose k items. choosing 0 items from the bag means insert your hand inside the bag and come up with empty hand, just to entertain the kid. You can still do that. So there is no 0 way of doing it. There is still 1 way of doing this.

share|cite|improve this answer

answered 10 hours ago

Neel Basu

19510

share|cite|improve this answer

share|cite|improve this answer

answered 10 hours ago

Neel Basu

19510

answered 10 hours ago

Neel Basu

19510

answered 10 hours ago

Neel Basu

19510

19510

add a comment | 

add a comment | 

up vote
1
down vote

Given you have an SO account, I'll add here a general technique for understanding combinatorics that continues to help me daily while doing intense low-level programming.

Bit-permutations can be sub-categorized into bit-combinations to make some counting principles more intuitive. Here's an example of classifying all 256 potential octets (8-bit bytes):

---

0-combinations (8-combinations via dual)

$vertunderline8choose0simeqoverline8choose8vert=underline1+overline1=textbf2$

(00000000) (11111111)

---

1-combinations (7-combinations via dual)

$vertunderline8choose1simeqoverline8choose7vert=underline8+overline8=textbf16$

(00000001) (11111110)
(00000010) (11111101)
(00000100) (11111011)
(00001000) (11110111)
(00010000) (11101111)
(00100000) (11011111)
(01000000) (10111111)
(10000000) (01111111)

---

2-combinations (6-combinations via dual)

$vertunderline8choose2simeqoverline8choose6vert=underline28+overline28=textbf56$

(00000011) (11111100)
(00000101) (11111010)
(00000110) (11111001)
(00001001) (11110110)
(00001010) (11110101)
(00001100) (11110011)
(00010001) (11101110)
(00010010) (11101101)
(00010100) (11101011)
(00011000) (11100111)
(00100001) (11011110)
(00100010) (11011101)
(00100100) (11011011)
(00101000) (11010111)
(00110000) (11001111)
(01000001) (10111110)
(01000010) (10111101)
(01000100) (10111011)
(01001000) (10110111)
(01010000) (10101111)
(01100000) (10011111)
(10000001) (01111110)
(10000010) (01111101)
(10000100) (01111011)
(10001000) (01110111)
(10010000) (01101111)
(10100000) (01011111)
(11000000) (00111111)

---

3-combinations (5-combinations via dual)

$vertunderline8choose3simeqoverline8choose5vert=underline56+overline56=textbf112$

(00000111) (11111000)
(00001011) (11110100)
(00001101) (11110010)
(00001110) (11110001)
(00010011) (11101100)
(00010101) (11101010)
(00010110) (11101001)
(00011001) (11100110)
(00011010) (11100101)
(00011100) (11100011)
(00100011) (11011100)
(00100101) (11011010)
(00100110) (11011001)
(00101001) (11010110)
(00101010) (11010101)
(00101100) (11010011)
(00110001) (11001110)
(00110010) (11001101)
(00110100) (11001011)
(00111000) (11000111)
(01000011) (10111100)
(01000101) (10111010)
(01000110) (10111001)
(01001001) (10110110)
(01001010) (10110101)
(01001100) (10110011)
(01010001) (10101110)
(01010010) (10101101)
(01010100) (10101011)
(01011000) (10100111)
(01100001) (10011110)
(01100010) (10011101)
(01100100) (10011011)
(01101000) (10010111)
(01110000) (10001111)
(10000011) (01111100)
(10000101) (01111010)
(10000110) (01111001)
(10001001) (01110110)
(10001010) (01110101)
(10001100) (01110011)
(10010001) (01101110)
(10010010) (01101101)
(10010100) (01101011)
(10011000) (01100111)
(10100001) (01011110)
(10100010) (01011101)
(10100100) (01011011)
(10101000) (01010111)
(10110000) (01001111)
(11000001) (00111110)
(11000010) (00111101)
(11000100) (00111011)
(11001000) (00110111)
(11010000) (00101111)
(11100000) (00011111)

---

4-combinations (self-dual)

$vertunderlineoverline8choose4vert=underlineoverline70=textbf70$

(00001111)
(00010111)
(00011011)
(00011101)
(00011110)
(00100111)
(00101011)
(00101101)
(00101110)
(00110011)
(00110101)
(00110110)
(00111001)
(00111010)
(00111100)
(01000111)
(01001011)
(01001101)
(01001110)
(01010011)
(01010101)
(01010110)
(01011001)
(01011010)
(01011100)
(01100011)
(01100101)
(01100110)
(01101001)
(01101010)
(01101100)
(01110001)
(01110010)
(01110100)
(01111000)
(10000111)
(10001011)
(10001101)
(10001110)
(10010011)
(10010101)
(10010110)
(10011001)
(10011010)
(10011100)
(10100011)
(10100101)
(10100110)
(10101001)
(10101010)
(10101100)
(10110001)
(10110010)
(10110100)
(10111000)
(11000011)
(11000101)
(11000110)
(11001001)
(11001010)
(11001100)
(11010001)
(11010010)
(11010100)
(11011000)
(11100001)
(11100010)
(11100100)
(11101000)
(11110000)

---

$vertPermutationsvert = 2+16+56+112+70=textbf256$

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answered 12 hours ago

user13972

1111

New contributor

user13972 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

up vote
1
down vote

Given you have an SO account, I'll add here a general technique for understanding combinatorics that continues to help me daily while doing intense low-level programming.

Bit-permutations can be sub-categorized into bit-combinations to make some counting principles more intuitive. Here's an example of classifying all 256 potential octets (8-bit bytes):

---

0-combinations (8-combinations via dual)

$vertunderline8choose0simeqoverline8choose8vert=underline1+overline1=textbf2$

(00000000) (11111111)

---

1-combinations (7-combinations via dual)

$vertunderline8choose1simeqoverline8choose7vert=underline8+overline8=textbf16$

(00000001) (11111110)
(00000010) (11111101)
(00000100) (11111011)
(00001000) (11110111)
(00010000) (11101111)
(00100000) (11011111)
(01000000) (10111111)
(10000000) (01111111)

---

2-combinations (6-combinations via dual)

$vertunderline8choose2simeqoverline8choose6vert=underline28+overline28=textbf56$

(00000011) (11111100)
(00000101) (11111010)
(00000110) (11111001)
(00001001) (11110110)
(00001010) (11110101)
(00001100) (11110011)
(00010001) (11101110)
(00010010) (11101101)
(00010100) (11101011)
(00011000) (11100111)
(00100001) (11011110)
(00100010) (11011101)
(00100100) (11011011)
(00101000) (11010111)
(00110000) (11001111)
(01000001) (10111110)
(01000010) (10111101)
(01000100) (10111011)
(01001000) (10110111)
(01010000) (10101111)
(01100000) (10011111)
(10000001) (01111110)
(10000010) (01111101)
(10000100) (01111011)
(10001000) (01110111)
(10010000) (01101111)
(10100000) (01011111)
(11000000) (00111111)

---

3-combinations (5-combinations via dual)

$vertunderline8choose3simeqoverline8choose5vert=underline56+overline56=textbf112$

(00000111) (11111000)
(00001011) (11110100)
(00001101) (11110010)
(00001110) (11110001)
(00010011) (11101100)
(00010101) (11101010)
(00010110) (11101001)
(00011001) (11100110)
(00011010) (11100101)
(00011100) (11100011)
(00100011) (11011100)
(00100101) (11011010)
(00100110) (11011001)
(00101001) (11010110)
(00101010) (11010101)
(00101100) (11010011)
(00110001) (11001110)
(00110010) (11001101)
(00110100) (11001011)
(00111000) (11000111)
(01000011) (10111100)
(01000101) (10111010)
(01000110) (10111001)
(01001001) (10110110)
(01001010) (10110101)
(01001100) (10110011)
(01010001) (10101110)
(01010010) (10101101)
(01010100) (10101011)
(01011000) (10100111)
(01100001) (10011110)
(01100010) (10011101)
(01100100) (10011011)
(01101000) (10010111)
(01110000) (10001111)
(10000011) (01111100)
(10000101) (01111010)
(10000110) (01111001)
(10001001) (01110110)
(10001010) (01110101)
(10001100) (01110011)
(10010001) (01101110)
(10010010) (01101101)
(10010100) (01101011)
(10011000) (01100111)
(10100001) (01011110)
(10100010) (01011101)
(10100100) (01011011)
(10101000) (01010111)
(10110000) (01001111)
(11000001) (00111110)
(11000010) (00111101)
(11000100) (00111011)
(11001000) (00110111)
(11010000) (00101111)
(11100000) (00011111)

---

4-combinations (self-dual)

$vertunderlineoverline8choose4vert=underlineoverline70=textbf70$

(00001111)
(00010111)
(00011011)
(00011101)
(00011110)
(00100111)
(00101011)
(00101101)
(00101110)
(00110011)
(00110101)
(00110110)
(00111001)
(00111010)
(00111100)
(01000111)
(01001011)
(01001101)
(01001110)
(01010011)
(01010101)
(01010110)
(01011001)
(01011010)
(01011100)
(01100011)
(01100101)
(01100110)
(01101001)
(01101010)
(01101100)
(01110001)
(01110010)
(01110100)
(01111000)
(10000111)
(10001011)
(10001101)
(10001110)
(10010011)
(10010101)
(10010110)
(10011001)
(10011010)
(10011100)
(10100011)
(10100101)
(10100110)
(10101001)
(10101010)
(10101100)
(10110001)
(10110010)
(10110100)
(10111000)
(11000011)
(11000101)
(11000110)
(11001001)
(11001010)
(11001100)
(11010001)
(11010010)
(11010100)
(11011000)
(11100001)
(11100010)
(11100100)
(11101000)
(11110000)

---

$vertPermutationsvert = 2+16+56+112+70=textbf256$

share|cite|improve this answer

answered 12 hours ago

user13972

1111

New contributor

user13972 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

up vote
1
down vote

up vote
1
down vote

Given you have an SO account, I'll add here a general technique for understanding combinatorics that continues to help me daily while doing intense low-level programming.

Bit-permutations can be sub-categorized into bit-combinations to make some counting principles more intuitive. Here's an example of classifying all 256 potential octets (8-bit bytes):

---

0-combinations (8-combinations via dual)

$vertunderline8choose0simeqoverline8choose8vert=underline1+overline1=textbf2$

(00000000) (11111111)

---

1-combinations (7-combinations via dual)

$vertunderline8choose1simeqoverline8choose7vert=underline8+overline8=textbf16$

(00000001) (11111110)
(00000010) (11111101)
(00000100) (11111011)
(00001000) (11110111)
(00010000) (11101111)
(00100000) (11011111)
(01000000) (10111111)
(10000000) (01111111)

---

2-combinations (6-combinations via dual)

$vertunderline8choose2simeqoverline8choose6vert=underline28+overline28=textbf56$

(00000011) (11111100)
(00000101) (11111010)
(00000110) (11111001)
(00001001) (11110110)
(00001010) (11110101)
(00001100) (11110011)
(00010001) (11101110)
(00010010) (11101101)
(00010100) (11101011)
(00011000) (11100111)
(00100001) (11011110)
(00100010) (11011101)
(00100100) (11011011)
(00101000) (11010111)
(00110000) (11001111)
(01000001) (10111110)
(01000010) (10111101)
(01000100) (10111011)
(01001000) (10110111)
(01010000) (10101111)
(01100000) (10011111)
(10000001) (01111110)
(10000010) (01111101)
(10000100) (01111011)
(10001000) (01110111)
(10010000) (01101111)
(10100000) (01011111)
(11000000) (00111111)

---

3-combinations (5-combinations via dual)

$vertunderline8choose3simeqoverline8choose5vert=underline56+overline56=textbf112$

(00000111) (11111000)
(00001011) (11110100)
(00001101) (11110010)
(00001110) (11110001)
(00010011) (11101100)
(00010101) (11101010)
(00010110) (11101001)
(00011001) (11100110)
(00011010) (11100101)
(00011100) (11100011)
(00100011) (11011100)
(00100101) (11011010)
(00100110) (11011001)
(00101001) (11010110)
(00101010) (11010101)
(00101100) (11010011)
(00110001) (11001110)
(00110010) (11001101)
(00110100) (11001011)
(00111000) (11000111)
(01000011) (10111100)
(01000101) (10111010)
(01000110) (10111001)
(01001001) (10110110)
(01001010) (10110101)
(01001100) (10110011)
(01010001) (10101110)
(01010010) (10101101)
(01010100) (10101011)
(01011000) (10100111)
(01100001) (10011110)
(01100010) (10011101)
(01100100) (10011011)
(01101000) (10010111)
(01110000) (10001111)
(10000011) (01111100)
(10000101) (01111010)
(10000110) (01111001)
(10001001) (01110110)
(10001010) (01110101)
(10001100) (01110011)
(10010001) (01101110)
(10010010) (01101101)
(10010100) (01101011)
(10011000) (01100111)
(10100001) (01011110)
(10100010) (01011101)
(10100100) (01011011)
(10101000) (01010111)
(10110000) (01001111)
(11000001) (00111110)
(11000010) (00111101)
(11000100) (00111011)
(11001000) (00110111)
(11010000) (00101111)
(11100000) (00011111)

---

4-combinations (self-dual)

$vertunderlineoverline8choose4vert=underlineoverline70=textbf70$

(00001111)
(00010111)
(00011011)
(00011101)
(00011110)
(00100111)
(00101011)
(00101101)
(00101110)
(00110011)
(00110101)
(00110110)
(00111001)
(00111010)
(00111100)
(01000111)
(01001011)
(01001101)
(01001110)
(01010011)
(01010101)
(01010110)
(01011001)
(01011010)
(01011100)
(01100011)
(01100101)
(01100110)
(01101001)
(01101010)
(01101100)
(01110001)
(01110010)
(01110100)
(01111000)
(10000111)
(10001011)
(10001101)
(10001110)
(10010011)
(10010101)
(10010110)
(10011001)
(10011010)
(10011100)
(10100011)
(10100101)
(10100110)
(10101001)
(10101010)
(10101100)
(10110001)
(10110010)
(10110100)
(10111000)
(11000011)
(11000101)
(11000110)
(11001001)
(11001010)
(11001100)
(11010001)
(11010010)
(11010100)
(11011000)
(11100001)
(11100010)
(11100100)
(11101000)
(11110000)

---

$vertPermutationsvert = 2+16+56+112+70=textbf256$

share|cite|improve this answer

answered 12 hours ago

user13972

1111

New contributor

user13972 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Given you have an SO account, I'll add here a general technique for understanding combinatorics that continues to help me daily while doing intense low-level programming.

Bit-permutations can be sub-categorized into bit-combinations to make some counting principles more intuitive. Here's an example of classifying all 256 potential octets (8-bit bytes):

---

0-combinations (8-combinations via dual)

$vertunderline8choose0simeqoverline8choose8vert=underline1+overline1=textbf2$

(00000000) (11111111)

---

1-combinations (7-combinations via dual)

$vertunderline8choose1simeqoverline8choose7vert=underline8+overline8=textbf16$

(00000001) (11111110)
(00000010) (11111101)
(00000100) (11111011)
(00001000) (11110111)
(00010000) (11101111)
(00100000) (11011111)
(01000000) (10111111)
(10000000) (01111111)

---

2-combinations (6-combinations via dual)

$vertunderline8choose2simeqoverline8choose6vert=underline28+overline28=textbf56$

(00000011) (11111100)
(00000101) (11111010)
(00000110) (11111001)
(00001001) (11110110)
(00001010) (11110101)
(00001100) (11110011)
(00010001) (11101110)
(00010010) (11101101)
(00010100) (11101011)
(00011000) (11100111)
(00100001) (11011110)
(00100010) (11011101)
(00100100) (11011011)
(00101000) (11010111)
(00110000) (11001111)
(01000001) (10111110)
(01000010) (10111101)
(01000100) (10111011)
(01001000) (10110111)
(01010000) (10101111)
(01100000) (10011111)
(10000001) (01111110)
(10000010) (01111101)
(10000100) (01111011)
(10001000) (01110111)
(10010000) (01101111)
(10100000) (01011111)
(11000000) (00111111)

---

3-combinations (5-combinations via dual)

$vertunderline8choose3simeqoverline8choose5vert=underline56+overline56=textbf112$

(00000111) (11111000)
(00001011) (11110100)
(00001101) (11110010)
(00001110) (11110001)
(00010011) (11101100)
(00010101) (11101010)
(00010110) (11101001)
(00011001) (11100110)
(00011010) (11100101)
(00011100) (11100011)
(00100011) (11011100)
(00100101) (11011010)
(00100110) (11011001)
(00101001) (11010110)
(00101010) (11010101)
(00101100) (11010011)
(00110001) (11001110)
(00110010) (11001101)
(00110100) (11001011)
(00111000) (11000111)
(01000011) (10111100)
(01000101) (10111010)
(01000110) (10111001)
(01001001) (10110110)
(01001010) (10110101)
(01001100) (10110011)
(01010001) (10101110)
(01010010) (10101101)
(01010100) (10101011)
(01011000) (10100111)
(01100001) (10011110)
(01100010) (10011101)
(01100100) (10011011)
(01101000) (10010111)
(01110000) (10001111)
(10000011) (01111100)
(10000101) (01111010)
(10000110) (01111001)
(10001001) (01110110)
(10001010) (01110101)
(10001100) (01110011)
(10010001) (01101110)
(10010010) (01101101)
(10010100) (01101011)
(10011000) (01100111)
(10100001) (01011110)
(10100010) (01011101)
(10100100) (01011011)
(10101000) (01010111)
(10110000) (01001111)
(11000001) (00111110)
(11000010) (00111101)
(11000100) (00111011)
(11001000) (00110111)
(11010000) (00101111)
(11100000) (00011111)

---

4-combinations (self-dual)

$vertunderlineoverline8choose4vert=underlineoverline70=textbf70$

(00001111)
(00010111)
(00011011)
(00011101)
(00011110)
(00100111)
(00101011)
(00101101)
(00101110)
(00110011)
(00110101)
(00110110)
(00111001)
(00111010)
(00111100)
(01000111)
(01001011)
(01001101)
(01001110)
(01010011)
(01010101)
(01010110)
(01011001)
(01011010)
(01011100)
(01100011)
(01100101)
(01100110)
(01101001)
(01101010)
(01101100)
(01110001)
(01110010)
(01110100)
(01111000)
(10000111)
(10001011)
(10001101)
(10001110)
(10010011)
(10010101)
(10010110)
(10011001)
(10011010)
(10011100)
(10100011)
(10100101)
(10100110)
(10101001)
(10101010)
(10101100)
(10110001)
(10110010)
(10110100)
(10111000)
(11000011)
(11000101)
(11000110)
(11001001)
(11001010)
(11001100)
(11010001)
(11010010)
(11010100)
(11011000)
(11100001)
(11100010)
(11100100)
(11101000)
(11110000)

---

$vertPermutationsvert = 2+16+56+112+70=textbf256$

share|cite|improve this answer

answered 12 hours ago

user13972

1111

New contributor

user13972 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this answer

share|cite|improve this answer

answered 12 hours ago

user13972

1111

New contributor

user13972 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 12 hours ago

user13972

1111

answered 12 hours ago

user13972

1111

1111

New contributor

user13972 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

user13972 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

user13972 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

add a comment | 

up vote
1
down vote

Imaginate as an option:

1) Pick $0$ elements

2) Pick $1$ element

3) Pick $2$ elements

...

n+1) Pick $n$ elements

---

(EDIT)
Life examples:

I. Voting: You can choose between Candidate A, Candidate B or no one by not voting.

Note that choosing nobody is a choose as well (not on paper of course).

II. Shopping: You want to buy a specific type of bread (like: gluten free)- so if there is- you pick or if there is no- 'that's your default vallue'.

Your set here is $text0,1$ with the cardinality of $2$ results.

III. A school mate in 2005:

"As you see (Chris) I have ten fingers: 1,2,3...10.

Let's count now from zero to be sure: 0,1,2...9"

This puzzle is not the best sample (since he could start to counter by minus one for example) but the point was: Finger #$0$ is a finger as well .

share|cite|improve this answer

edited 12 hours ago

answered yesterday

Krzysztof Myśliwiec

51214

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Could you elaborate, please?
  – DaG
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DaG Yes- tommorrow I made an edit ( hopefully without english grammar mistakes).
  – Krzysztof Myśliwiec
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DaG I not unclude terms like the Empty Set to not complicate my primary answer
  – Krzysztof Myśliwiec
  12 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

Imaginate as an option:

1) Pick $0$ elements

2) Pick $1$ element

3) Pick $2$ elements

...

n+1) Pick $n$ elements

---

(EDIT)
Life examples:

I. Voting: You can choose between Candidate A, Candidate B or no one by not voting.

Note that choosing nobody is a choose as well (not on paper of course).

II. Shopping: You want to buy a specific type of bread (like: gluten free)- so if there is- you pick or if there is no- 'that's your default vallue'.

Your set here is $text0,1$ with the cardinality of $2$ results.

III. A school mate in 2005:

"As you see (Chris) I have ten fingers: 1,2,3...10.

Let's count now from zero to be sure: 0,1,2...9"

This puzzle is not the best sample (since he could start to counter by minus one for example) but the point was: Finger #$0$ is a finger as well .

share|cite|improve this answer

edited 12 hours ago

answered yesterday

Krzysztof Myśliwiec

51214

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Could you elaborate, please?
  – DaG
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DaG Yes- tommorrow I made an edit ( hopefully without english grammar mistakes).
  – Krzysztof Myśliwiec
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DaG I not unclude terms like the Empty Set to not complicate my primary answer
  – Krzysztof Myśliwiec
  12 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Imaginate as an option:

1) Pick $0$ elements

2) Pick $1$ element

3) Pick $2$ elements

...

n+1) Pick $n$ elements

---

(EDIT)
Life examples:

I. Voting: You can choose between Candidate A, Candidate B or no one by not voting.

Note that choosing nobody is a choose as well (not on paper of course).

II. Shopping: You want to buy a specific type of bread (like: gluten free)- so if there is- you pick or if there is no- 'that's your default vallue'.

Your set here is $text0,1$ with the cardinality of $2$ results.

III. A school mate in 2005:

"As you see (Chris) I have ten fingers: 1,2,3...10.

Let's count now from zero to be sure: 0,1,2...9"

This puzzle is not the best sample (since he could start to counter by minus one for example) but the point was: Finger #$0$ is a finger as well .

share|cite|improve this answer

edited 12 hours ago

answered yesterday

Krzysztof Myśliwiec

51214

Imaginate as an option:

1) Pick $0$ elements

2) Pick $1$ element

3) Pick $2$ elements

...

n+1) Pick $n$ elements

---

(EDIT)
Life examples:

I. Voting: You can choose between Candidate A, Candidate B or no one by not voting.

Note that choosing nobody is a choose as well (not on paper of course).

II. Shopping: You want to buy a specific type of bread (like: gluten free)- so if there is- you pick or if there is no- 'that's your default vallue'.

Your set here is $text0,1$ with the cardinality of $2$ results.

III. A school mate in 2005:

"As you see (Chris) I have ten fingers: 1,2,3...10.

Let's count now from zero to be sure: 0,1,2...9"

This puzzle is not the best sample (since he could start to counter by minus one for example) but the point was: Finger #$0$ is a finger as well .

share|cite|improve this answer

edited 12 hours ago

answered yesterday

Krzysztof Myśliwiec

51214

share|cite|improve this answer

share|cite|improve this answer

edited 12 hours ago

edited 12 hours ago

edited 12 hours ago

answered yesterday

Krzysztof Myśliwiec

51214

answered yesterday

Krzysztof Myśliwiec

51214

answered yesterday

Krzysztof Myśliwiec

51214

51214

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Could you elaborate, please?
  – DaG
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DaG Yes- tommorrow I made an edit ( hopefully without english grammar mistakes).
  – Krzysztof Myśliwiec
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DaG I not unclude terms like the Empty Set to not complicate my primary answer
  – Krzysztof Myśliwiec
  12 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Could you elaborate, please?
  – DaG
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DaG Yes- tommorrow I made an edit ( hopefully without english grammar mistakes).
  – Krzysztof Myśliwiec
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DaG I not unclude terms like the Empty Set to not complicate my primary answer
  – Krzysztof Myśliwiec
  12 hours ago
  

  
  
  
  

3

3

Could you elaborate, please?
– DaG
yesterday

Could you elaborate, please?
– DaG
yesterday

@DaG Yes- tommorrow I made an edit ( hopefully without english grammar mistakes).
– Krzysztof Myśliwiec
yesterday

@DaG Yes- tommorrow I made an edit ( hopefully without english grammar mistakes).
– Krzysztof Myśliwiec
yesterday

@DaG I not unclude terms like the Empty Set to not complicate my primary answer
– Krzysztof Myśliwiec
12 hours ago

@DaG I not unclude terms like the Empty Set to not complicate my primary answer
– Krzysztof Myśliwiec
12 hours ago

add a comment | 

up vote
1
down vote

Think of it like this: Choose the items that you don't want to choose. By choosing $k$ items out of $n$ not to choose, you will effectively have chosen $n - k$ out of the $n$ items, and it becomes obvious that there are equally many ways of choosing $k$ items out of $n$ as there are ways of choosing $n-k$ items out of $n$.

So, there are equally many ways of choosing 0 items out of $n$ as there are ways of choosing $n$ items out of $n$.

share|cite|improve this answer

answered 10 hours ago

HelloGoodbye

1905

add a comment | 

up vote
1
down vote

Think of it like this: Choose the items that you don't want to choose. By choosing $k$ items out of $n$ not to choose, you will effectively have chosen $n - k$ out of the $n$ items, and it becomes obvious that there are equally many ways of choosing $k$ items out of $n$ as there are ways of choosing $n-k$ items out of $n$.

So, there are equally many ways of choosing 0 items out of $n$ as there are ways of choosing $n$ items out of $n$.

share|cite|improve this answer

answered 10 hours ago

HelloGoodbye

1905

add a comment | 

up vote
1
down vote

up vote
1
down vote

Think of it like this: Choose the items that you don't want to choose. By choosing $k$ items out of $n$ not to choose, you will effectively have chosen $n - k$ out of the $n$ items, and it becomes obvious that there are equally many ways of choosing $k$ items out of $n$ as there are ways of choosing $n-k$ items out of $n$.

So, there are equally many ways of choosing 0 items out of $n$ as there are ways of choosing $n$ items out of $n$.

share|cite|improve this answer

answered 10 hours ago

HelloGoodbye

1905

Think of it like this: Choose the items that you don't want to choose. By choosing $k$ items out of $n$ not to choose, you will effectively have chosen $n - k$ out of the $n$ items, and it becomes obvious that there are equally many ways of choosing $k$ items out of $n$ as there are ways of choosing $n-k$ items out of $n$.

So, there are equally many ways of choosing 0 items out of $n$ as there are ways of choosing $n$ items out of $n$.

share|cite|improve this answer

answered 10 hours ago

HelloGoodbye

1905

share|cite|improve this answer

share|cite|improve this answer

answered 10 hours ago

HelloGoodbye

1905

answered 10 hours ago

HelloGoodbye

1905

answered 10 hours ago

HelloGoodbye

1905

1905

add a comment | 

add a comment | 

up vote
1
down vote

The other answers are fine, but for a geometric twist:

Pascal's triangle would be a lot less elegant without this definition:

 1
 1 1
 1 2 1
 1 3 3 1
1 4 6 4 1

(And so on.) Even if there wasn't any other compelling reason for defining $n choose k$ in such a way that $nchoose 0 = 0$, the symmetry of the triangle itself could lead you there. It would be odd if the last 4 entries of the bottom row were $4choose 1, ldots, 4 choose 4$ but the first entry wasn't $4 choose 0$.

share|cite|improve this answer

answered 3 hours ago

John Coleman

3,66811123

add a comment | 

up vote
1
down vote

The other answers are fine, but for a geometric twist:

Pascal's triangle would be a lot less elegant without this definition:

 1
 1 1
 1 2 1
 1 3 3 1
1 4 6 4 1

(And so on.) Even if there wasn't any other compelling reason for defining $n choose k$ in such a way that $nchoose 0 = 0$, the symmetry of the triangle itself could lead you there. It would be odd if the last 4 entries of the bottom row were $4choose 1, ldots, 4 choose 4$ but the first entry wasn't $4 choose 0$.

share|cite|improve this answer

answered 3 hours ago

John Coleman

3,66811123

add a comment | 

up vote
1
down vote

up vote
1
down vote

The other answers are fine, but for a geometric twist:

Pascal's triangle would be a lot less elegant without this definition:

 1
 1 1
 1 2 1
 1 3 3 1
1 4 6 4 1

(And so on.) Even if there wasn't any other compelling reason for defining $n choose k$ in such a way that $nchoose 0 = 0$, the symmetry of the triangle itself could lead you there. It would be odd if the last 4 entries of the bottom row were $4choose 1, ldots, 4 choose 4$ but the first entry wasn't $4 choose 0$.

share|cite|improve this answer

answered 3 hours ago

John Coleman

3,66811123

The other answers are fine, but for a geometric twist:

Pascal's triangle would be a lot less elegant without this definition:

 1
 1 1
 1 2 1
 1 3 3 1
1 4 6 4 1

(And so on.) Even if there wasn't any other compelling reason for defining $n choose k$ in such a way that $nchoose 0 = 0$, the symmetry of the triangle itself could lead you there. It would be odd if the last 4 entries of the bottom row were $4choose 1, ldots, 4 choose 4$ but the first entry wasn't $4 choose 0$.

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answered 3 hours ago

John Coleman

3,66811123

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answered 3 hours ago

John Coleman

3,66811123

answered 3 hours ago

John Coleman

3,66811123

answered 3 hours ago

John Coleman

3,66811123

3,66811123

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up vote
1
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How to choose 0 items: Choose n items, then pick the ones that remain. There is one way to choose n items, therefore there is only one way to pick the ones that remain.

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answered 2 hours ago

gnasher729

5,9111028

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up vote
1
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How to choose 0 items: Choose n items, then pick the ones that remain. There is one way to choose n items, therefore there is only one way to pick the ones that remain.

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answered 2 hours ago

gnasher729

5,9111028

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up vote
1
down vote

up vote
1
down vote

How to choose 0 items: Choose n items, then pick the ones that remain. There is one way to choose n items, therefore there is only one way to pick the ones that remain.

share|cite|improve this answer

answered 2 hours ago

gnasher729

5,9111028

How to choose 0 items: Choose n items, then pick the ones that remain. There is one way to choose n items, therefore there is only one way to pick the ones that remain.

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answered 2 hours ago

gnasher729

5,9111028

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share|cite|improve this answer

answered 2 hours ago

gnasher729

5,9111028

answered 2 hours ago

gnasher729

5,9111028

answered 2 hours ago

gnasher729

5,9111028

5,9111028

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subba is a new contributor. Be nice, and check out our Code of Conduct.

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