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Why is 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
asked 3 hours ago
subba
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up vote
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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
asked 3 hours ago
subba
1133
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subba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Is it possible to choose none of the $n$ items? Is there more than one way to do that?
– bof
3 hours ago
2
The only choice you can make is $lbrace rbrace$
– P. Quinton
3 hours ago
What do you think it should be?
– Martin R
3 hours ago
The number of bitstrings of length $n$ containing $n$ ones is the same as the number containing $n$ zeros.
– bof
3 hours ago
If each of the $n$ items may be chosen or refused, why would the number of ways of choosing $n$ items be different from the number of ways of refusing $n$ items?
– bof
3 hours ago
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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
asked 3 hours ago
subba
1133
New contributor
subba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
asked 3 hours ago
subba
1133
New contributor
subba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
subba
1133
New contributor
subba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
subba
1133
New contributor
subba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
subba
1133
asked 3 hours ago
subba
1133
1133
New contributor
subba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
subba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
subba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Is it possible to choose none of the $n$ items? Is there more than one way to do that?
– bof
3 hours ago
2
The only choice you can make is $lbrace rbrace$
– P. Quinton
3 hours ago
What do you think it should be?
– Martin R
3 hours ago
The number of bitstrings of length $n$ containing $n$ ones is the same as the number containing $n$ zeros.
– bof
3 hours ago
If each of the $n$ items may be chosen or refused, why would the number of ways of choosing $n$ items be different from the number of ways of refusing $n$ items?
– bof
3 hours ago
 |Â
show 1 more comment
Is it possible to choose none of the $n$ items? Is there more than one way to do that?
– bof
3 hours ago
2
The only choice you can make is $lbrace rbrace$
– P. Quinton
3 hours ago
What do you think it should be?
– Martin R
3 hours ago
The number of bitstrings of length $n$ containing $n$ ones is the same as the number containing $n$ zeros.
– bof
3 hours ago
If each of the $n$ items may be chosen or refused, why would the number of ways of choosing $n$ items be different from the number of ways of refusing $n$ items?
– bof
3 hours ago
Is it possible to choose none of the $n$ items? Is there more than one way to do that?
– bof
3 hours ago
Is it possible to choose none of the $n$ items? Is there more than one way to do that?
– bof
3 hours ago
2
2
The only choice you can make is $lbrace rbrace$
– P. Quinton
3 hours ago
The only choice you can make is $lbrace rbrace$
– P. Quinton
3 hours ago
What do you think it should be?
– Martin R
3 hours ago
What do you think it should be?
– Martin R
3 hours ago
The number of bitstrings of length $n$ containing $n$ ones is the same as the number containing $n$ zeros.
– bof
3 hours ago
The number of bitstrings of length $n$ containing $n$ ones is the same as the number containing $n$ zeros.
– bof
3 hours ago
If each of the $n$ items may be chosen or refused, why would the number of ways of choosing $n$ items be different from the number of ways of refusing $n$ items?
– bof
3 hours ago
If each of the $n$ items may be chosen or refused, why would the number of ways of choosing $n$ items be different from the number of ways of refusing $n$ items?
– bof
3 hours ago
 |Â
show 1 more comment
3 Answers
3
active
oldest
votes
up vote
3
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:
- The number of subsets of an $n$-element set with $0$ elements (and we always assume the empty set counts)
- 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).
- 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.
answered 2 hours ago
Kevin Carlson
30.7k23068
add a comment |Â
up vote
0
down vote
Imaginate as an option:
1) Pick $0$ elements
2) Pick $1$ element
3) Pick $2$ elements
...
n+1) Pick $n$ elements
answered 3 hours ago
Krzysztof Myśliwiec
33214
add a comment |Â
up vote
0
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 $emptyset$.
answered 3 hours ago
ArsenBerk
7,39121234
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
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:
- The number of subsets of an $n$-element set with $0$ elements (and we always assume the empty set counts)
- 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).
- 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.
answered 2 hours ago
Kevin Carlson
30.7k23068
add a comment |Â
up vote
3
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:
- The number of subsets of an $n$-element set with $0$ elements (and we always assume the empty set counts)
- 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).
- 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.
answered 2 hours ago
Kevin Carlson
30.7k23068
add a comment |Â
up vote
3
down vote
accepted
up vote
3
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:
- The number of subsets of an $n$-element set with $0$ elements (and we always assume the empty set counts)
- 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).
- 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.
answered 2 hours ago
Kevin Carlson
30.7k23068
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:
- The number of subsets of an $n$-element set with $0$ elements (and we always assume the empty set counts)
- 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).
- 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.
answered 2 hours ago
Kevin Carlson
30.7k23068
answered 2 hours ago
Kevin Carlson
30.7k23068
answered 2 hours ago
Kevin Carlson
30.7k23068
answered 2 hours ago
Kevin Carlson
30.7k23068
30.7k23068
add a comment |Â
add a comment |Â
up vote
0
down vote
Imaginate as an option:
1) Pick $0$ elements
2) Pick $1$ element
3) Pick $2$ elements
...
n+1) Pick $n$ elements
answered 3 hours ago
Krzysztof Myśliwiec
33214
add a comment |Â
up vote
0
down vote
Imaginate as an option:
1) Pick $0$ elements
2) Pick $1$ element
3) Pick $2$ elements
...
n+1) Pick $n$ elements
answered 3 hours ago
Krzysztof Myśliwiec
33214
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Imaginate as an option:
1) Pick $0$ elements
2) Pick $1$ element
3) Pick $2$ elements
...
n+1) Pick $n$ elements
answered 3 hours ago
Krzysztof Myśliwiec
33214
Imaginate as an option:
1) Pick $0$ elements
2) Pick $1$ element
3) Pick $2$ elements
...
n+1) Pick $n$ elements
answered 3 hours ago
Krzysztof Myśliwiec
33214
answered 3 hours ago
Krzysztof Myśliwiec
33214
answered 3 hours ago
Krzysztof Myśliwiec
33214
answered 3 hours ago
Krzysztof Myśliwiec
33214
33214
add a comment |Â
add a comment |Â
up vote
0
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 $emptyset$.
answered 3 hours ago
ArsenBerk
7,39121234
add a comment |Â
up vote
0
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 $emptyset$.
answered 3 hours ago
ArsenBerk
7,39121234
add a comment |Â
up vote
0
down vote
up vote
0
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 $emptyset$.
answered 3 hours ago
ArsenBerk
7,39121234
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 $emptyset$.
answered 3 hours ago
ArsenBerk
7,39121234
answered 3 hours ago
ArsenBerk
7,39121234
answered 3 hours ago
ArsenBerk
7,39121234
answered 3 hours ago
ArsenBerk
7,39121234
7,39121234
add a comment |Â
add a comment |Â
subba is a new contributor. Be nice, and check out our Code of Conduct.
subba is a new contributor. Be nice, and check out our Code of Conduct.
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Is it possible to choose none of the $n$ items? Is there more than one way to do that?
– bof
3 hours ago
2
The only choice you can make is $lbrace rbrace$
– P. Quinton
3 hours ago
What do you think it should be?
– Martin R
3 hours ago
The number of bitstrings of length $n$ containing $n$ ones is the same as the number containing $n$ zeros.
– bof
3 hours ago
If each of the $n$ items may be chosen or refused, why would the number of ways of choosing $n$ items be different from the number of ways of refusing $n$ items?
– bof
3 hours ago