Mixing
Different ways of counting the same combination
Clash Royale CLAN TAG#URR8PPP
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What’s wrong with this reasoning:
How many different ways to pick a team of 3 from 4 people? $4 choose 3$
Alternative way of counting is to choose a team of 2 from the 4 first ($4 choose 2$ ways), and then choose a third person (2 people remaining so 2 ways). Then, since order doesn’t matter (i.e., we could have chosen the single person first, and then the team of two) so divide by $2!$. But $4 choose 2 times frac22! neq 4 choose 3$.
combinatorics
edited 4 mins ago
Ben
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asked 3 hours ago
Basil Beirouti
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up vote
2
down vote
favorite
What’s wrong with this reasoning:
How many different ways to pick a team of 3 from 4 people? $4 choose 3$
Alternative way of counting is to choose a team of 2 from the 4 first ($4 choose 2$ ways), and then choose a third person (2 people remaining so 2 ways). Then, since order doesn’t matter (i.e., we could have chosen the single person first, and then the team of two) so divide by $2!$. But $4 choose 2 times frac22! neq 4 choose 3$.
combinatorics
edited 4 mins ago
Ben
15.4k12181
asked 3 hours ago
Basil Beirouti
111
New contributor
Basil Beirouti 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 |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
What’s wrong with this reasoning:
How many different ways to pick a team of 3 from 4 people? $4 choose 3$
Alternative way of counting is to choose a team of 2 from the 4 first ($4 choose 2$ ways), and then choose a third person (2 people remaining so 2 ways). Then, since order doesn’t matter (i.e., we could have chosen the single person first, and then the team of two) so divide by $2!$. But $4 choose 2 times frac22! neq 4 choose 3$.
combinatorics
edited 4 mins ago
Ben
15.4k12181
asked 3 hours ago
Basil Beirouti
111
New contributor
Basil Beirouti is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What’s wrong with this reasoning:
How many different ways to pick a team of 3 from 4 people? $4 choose 3$
Alternative way of counting is to choose a team of 2 from the 4 first ($4 choose 2$ ways), and then choose a third person (2 people remaining so 2 ways). Then, since order doesn’t matter (i.e., we could have chosen the single person first, and then the team of two) so divide by $2!$. But $4 choose 2 times frac22! neq 4 choose 3$.
combinatorics
combinatorics
edited 4 mins ago
Ben
15.4k12181
asked 3 hours ago
Basil Beirouti
111
New contributor
Basil Beirouti is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 mins ago
Ben
15.4k12181
asked 3 hours ago
Basil Beirouti
111
New contributor
Basil Beirouti is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 mins ago
Ben
15.4k12181
edited 4 mins ago
Ben
15.4k12181
edited 4 mins ago
Ben
15.4k12181
15.4k12181
asked 3 hours ago
Basil Beirouti
111
New contributor
Basil Beirouti 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
Basil Beirouti
111
asked 3 hours ago
Basil Beirouti
111
111
New contributor
Basil Beirouti is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Basil Beirouti is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Basil Beirouti 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 |Â
add a comment |Â
1 Answer
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It's ok to think about choosing 2 people first, then a third person. But, care must be taken to avoid double counting. For example, suppose we have people $A, B, C, D$. We pick $A$ and $B$ as the initial team of two, leaving $C$ and $D$ as the remaining choices. This gives possibilities: $A, B, C$ and $A, B, D$ for our 3-person team. But, note that each of these teams could also have resulted from a different pick of the initial two person team. For example, $A,B,C$ would also be produced by picking $A$ and $C$ followed by $B$, or $B$ and $C$ followed by $A$.
In general, suppose we have $n$ people and want to pick a $k$ person team. We do this by first picking $c < k$ people. There are $binomnc$ to do this. This leaves $n-c$ people, from which we must pick $k-c$ people to complete the team. There are $binomn-ck-c$ ways to do this. Multiplying these numbers, we have $binomnc binomn-ck-c$ possibilities. But, as above, many of these possibilities will be identical, and we must avoid counting them multiple times. The number of times a given set of $k$ people appear in our list of possible teams is $binomkc$, because this is the number of ways that those particular people could have been picked according to our procedure (i.e. by first picking $c$ of them). So, we divide by this number to obtain the number of unique possible teams:
$$fracbinomnc binomn-ck-cbinomkc$$
Writing out the expression shows that it's equivalent to $binomnk$.
answered 2 hours ago
user20160
14.6k12451
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
It's ok to think about choosing 2 people first, then a third person. But, care must be taken to avoid double counting. For example, suppose we have people $A, B, C, D$. We pick $A$ and $B$ as the initial team of two, leaving $C$ and $D$ as the remaining choices. This gives possibilities: $A, B, C$ and $A, B, D$ for our 3-person team. But, note that each of these teams could also have resulted from a different pick of the initial two person team. For example, $A,B,C$ would also be produced by picking $A$ and $C$ followed by $B$, or $B$ and $C$ followed by $A$.
In general, suppose we have $n$ people and want to pick a $k$ person team. We do this by first picking $c < k$ people. There are $binomnc$ to do this. This leaves $n-c$ people, from which we must pick $k-c$ people to complete the team. There are $binomn-ck-c$ ways to do this. Multiplying these numbers, we have $binomnc binomn-ck-c$ possibilities. But, as above, many of these possibilities will be identical, and we must avoid counting them multiple times. The number of times a given set of $k$ people appear in our list of possible teams is $binomkc$, because this is the number of ways that those particular people could have been picked according to our procedure (i.e. by first picking $c$ of them). So, we divide by this number to obtain the number of unique possible teams:
$$fracbinomnc binomn-ck-cbinomkc$$
Writing out the expression shows that it's equivalent to $binomnk$.
answered 2 hours ago
user20160
14.6k12451
add a comment |Â
up vote
2
down vote
It's ok to think about choosing 2 people first, then a third person. But, care must be taken to avoid double counting. For example, suppose we have people $A, B, C, D$. We pick $A$ and $B$ as the initial team of two, leaving $C$ and $D$ as the remaining choices. This gives possibilities: $A, B, C$ and $A, B, D$ for our 3-person team. But, note that each of these teams could also have resulted from a different pick of the initial two person team. For example, $A,B,C$ would also be produced by picking $A$ and $C$ followed by $B$, or $B$ and $C$ followed by $A$.
In general, suppose we have $n$ people and want to pick a $k$ person team. We do this by first picking $c < k$ people. There are $binomnc$ to do this. This leaves $n-c$ people, from which we must pick $k-c$ people to complete the team. There are $binomn-ck-c$ ways to do this. Multiplying these numbers, we have $binomnc binomn-ck-c$ possibilities. But, as above, many of these possibilities will be identical, and we must avoid counting them multiple times. The number of times a given set of $k$ people appear in our list of possible teams is $binomkc$, because this is the number of ways that those particular people could have been picked according to our procedure (i.e. by first picking $c$ of them). So, we divide by this number to obtain the number of unique possible teams:
$$fracbinomnc binomn-ck-cbinomkc$$
Writing out the expression shows that it's equivalent to $binomnk$.
answered 2 hours ago
user20160
14.6k12451
add a comment |Â
up vote
2
down vote
up vote
2
down vote
It's ok to think about choosing 2 people first, then a third person. But, care must be taken to avoid double counting. For example, suppose we have people $A, B, C, D$. We pick $A$ and $B$ as the initial team of two, leaving $C$ and $D$ as the remaining choices. This gives possibilities: $A, B, C$ and $A, B, D$ for our 3-person team. But, note that each of these teams could also have resulted from a different pick of the initial two person team. For example, $A,B,C$ would also be produced by picking $A$ and $C$ followed by $B$, or $B$ and $C$ followed by $A$.
In general, suppose we have $n$ people and want to pick a $k$ person team. We do this by first picking $c < k$ people. There are $binomnc$ to do this. This leaves $n-c$ people, from which we must pick $k-c$ people to complete the team. There are $binomn-ck-c$ ways to do this. Multiplying these numbers, we have $binomnc binomn-ck-c$ possibilities. But, as above, many of these possibilities will be identical, and we must avoid counting them multiple times. The number of times a given set of $k$ people appear in our list of possible teams is $binomkc$, because this is the number of ways that those particular people could have been picked according to our procedure (i.e. by first picking $c$ of them). So, we divide by this number to obtain the number of unique possible teams:
$$fracbinomnc binomn-ck-cbinomkc$$
Writing out the expression shows that it's equivalent to $binomnk$.
answered 2 hours ago
user20160
14.6k12451
It's ok to think about choosing 2 people first, then a third person. But, care must be taken to avoid double counting. For example, suppose we have people $A, B, C, D$. We pick $A$ and $B$ as the initial team of two, leaving $C$ and $D$ as the remaining choices. This gives possibilities: $A, B, C$ and $A, B, D$ for our 3-person team. But, note that each of these teams could also have resulted from a different pick of the initial two person team. For example, $A,B,C$ would also be produced by picking $A$ and $C$ followed by $B$, or $B$ and $C$ followed by $A$.
In general, suppose we have $n$ people and want to pick a $k$ person team. We do this by first picking $c < k$ people. There are $binomnc$ to do this. This leaves $n-c$ people, from which we must pick $k-c$ people to complete the team. There are $binomn-ck-c$ ways to do this. Multiplying these numbers, we have $binomnc binomn-ck-c$ possibilities. But, as above, many of these possibilities will be identical, and we must avoid counting them multiple times. The number of times a given set of $k$ people appear in our list of possible teams is $binomkc$, because this is the number of ways that those particular people could have been picked according to our procedure (i.e. by first picking $c$ of them). So, we divide by this number to obtain the number of unique possible teams:
$$fracbinomnc binomn-ck-cbinomkc$$
Writing out the expression shows that it's equivalent to $binomnk$.
answered 2 hours ago
user20160
14.6k12451
answered 2 hours ago
user20160
14.6k12451
answered 2 hours ago
user20160
14.6k12451
answered 2 hours ago
user20160
14.6k12451
14.6k12451
add a comment |Â
add a comment |Â
Basil Beirouti is a new contributor. Be nice, and check out our Code of Conduct.
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Basil Beirouti is a new contributor. Be nice, and check out our Code of Conduct.
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