Mixing
Combination with partial constraint
$begingroup$
We have a group of 15 people and want to find out how many possibilities there are to choose a subgroup of 5 people. This is pretty easy, we can just calculate $15 choose 5$.
But now suppose we have a constraint where two people don't want to be in the subgroup together. My first guess would have been to calculate how many possibilities there are that these two are in the same subgroup (which would be $15 choose 3$, right?) and then subtract that, so my solution to this problem would look like this:
${15 choose 5} - {15 choose 3} = 3003 - 455 = 2548$
Is this the correct way to approach this problem? If now, what would be the right answer?
combinatorics
asked Jan 20 at 11:06
Simon SchillerSimon Schiller
1033
$endgroup$
add a comment |
$begingroup$
We have a group of 15 people and want to find out how many possibilities there are to choose a subgroup of 5 people. This is pretty easy, we can just calculate $15 choose 5$.
But now suppose we have a constraint where two people don't want to be in the subgroup together. My first guess would have been to calculate how many possibilities there are that these two are in the same subgroup (which would be $15 choose 3$, right?) and then subtract that, so my solution to this problem would look like this:
${15 choose 5} - {15 choose 3} = 3003 - 455 = 2548$
Is this the correct way to approach this problem? If now, what would be the right answer?
combinatorics
asked Jan 20 at 11:06
Simon SchillerSimon Schiller
1033
$endgroup$
$begingroup$
The negation of two people being in the subgroup together is that at least one of them is not in the subgroup, not that neither of them are in the subgroup
$endgroup$
– user289143
Jan 20 at 11:21
add a comment |
$begingroup$
We have a group of 15 people and want to find out how many possibilities there are to choose a subgroup of 5 people. This is pretty easy, we can just calculate $15 choose 5$.
But now suppose we have a constraint where two people don't want to be in the subgroup together. My first guess would have been to calculate how many possibilities there are that these two are in the same subgroup (which would be $15 choose 3$, right?) and then subtract that, so my solution to this problem would look like this:
${15 choose 5} - {15 choose 3} = 3003 - 455 = 2548$
Is this the correct way to approach this problem? If now, what would be the right answer?
combinatorics
asked Jan 20 at 11:06
Simon SchillerSimon Schiller
1033
$endgroup$
We have a group of 15 people and want to find out how many possibilities there are to choose a subgroup of 5 people. This is pretty easy, we can just calculate $15 choose 5$.
But now suppose we have a constraint where two people don't want to be in the subgroup together. My first guess would have been to calculate how many possibilities there are that these two are in the same subgroup (which would be $15 choose 3$, right?) and then subtract that, so my solution to this problem would look like this:
${15 choose 5} - {15 choose 3} = 3003 - 455 = 2548$
Is this the correct way to approach this problem? If now, what would be the right answer?
combinatorics
combinatorics
asked Jan 20 at 11:06
Simon SchillerSimon Schiller
1033
asked Jan 20 at 11:06
Simon SchillerSimon Schiller
1033
asked Jan 20 at 11:06
Simon SchillerSimon Schiller
1033
asked Jan 20 at 11:06
Simon SchillerSimon Schiller
1033
asked Jan 20 at 11:06
Simon SchillerSimon Schiller
1033
1033
$begingroup$
The negation of two people being in the subgroup together is that at least one of them is not in the subgroup, not that neither of them are in the subgroup
$endgroup$
– user289143
Jan 20 at 11:21
add a comment |
$begingroup$
The negation of two people being in the subgroup together is that at least one of them is not in the subgroup, not that neither of them are in the subgroup
$endgroup$
– user289143
Jan 20 at 11:21
$begingroup$
The negation of two people being in the subgroup together is that at least one of them is not in the subgroup, not that neither of them are in the subgroup
$endgroup$
– user289143
Jan 20 at 11:21
$begingroup$
The negation of two people being in the subgroup together is that at least one of them is not in the subgroup, not that neither of them are in the subgroup
$endgroup$
– user289143
Jan 20 at 11:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There are three possible types of subgroups:
a) None of the two persons are in the subgroup
b) Only the first one is in the subgroup
c) Only the second one is in the subgroup
For a) is the same to choose $5$ people out of $13$, i.e. ${13}choose{5}$
For b) is the same to choose $4$ people out of $13$ (because you have already one and you don't want the other), i.e. ${13}choose {4}$
For c) is the same as b)
Thus the solution is ${13}choose {5}$ +$2 $${13}choose{4}$ $=1287+2 cdot 715=1287+1430=2717$
answered Jan 20 at 11:13
user289143user289143
1,002313
$endgroup$
$begingroup$
Okay, so my mistake was that I calculated $15 choose 3$, I should have done $13 choose 3$ because I already have two people. Then I get to the same result as you did. So ${15 choose 5} - {13 choose 3}$ works out as well. Thank you very much.
$endgroup$
– Simon Schiller
Jan 20 at 11:21
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are three possible types of subgroups:
a) None of the two persons are in the subgroup
b) Only the first one is in the subgroup
c) Only the second one is in the subgroup
For a) is the same to choose $5$ people out of $13$, i.e. ${13}choose{5}$
For b) is the same to choose $4$ people out of $13$ (because you have already one and you don't want the other), i.e. ${13}choose {4}$
For c) is the same as b)
Thus the solution is ${13}choose {5}$ +$2 $${13}choose{4}$ $=1287+2 cdot 715=1287+1430=2717$
answered Jan 20 at 11:13
user289143user289143
1,002313
$endgroup$
$begingroup$
Okay, so my mistake was that I calculated $15 choose 3$, I should have done $13 choose 3$ because I already have two people. Then I get to the same result as you did. So ${15 choose 5} - {13 choose 3}$ works out as well. Thank you very much.
$endgroup$
– Simon Schiller
Jan 20 at 11:21
add a comment |
$begingroup$
There are three possible types of subgroups:
a) None of the two persons are in the subgroup
b) Only the first one is in the subgroup
c) Only the second one is in the subgroup
For a) is the same to choose $5$ people out of $13$, i.e. ${13}choose{5}$
For b) is the same to choose $4$ people out of $13$ (because you have already one and you don't want the other), i.e. ${13}choose {4}$
For c) is the same as b)
Thus the solution is ${13}choose {5}$ +$2 $${13}choose{4}$ $=1287+2 cdot 715=1287+1430=2717$
answered Jan 20 at 11:13
user289143user289143
1,002313
$endgroup$
$begingroup$
Okay, so my mistake was that I calculated $15 choose 3$, I should have done $13 choose 3$ because I already have two people. Then I get to the same result as you did. So ${15 choose 5} - {13 choose 3}$ works out as well. Thank you very much.
$endgroup$
– Simon Schiller
Jan 20 at 11:21
add a comment |
$begingroup$
There are three possible types of subgroups:
a) None of the two persons are in the subgroup
b) Only the first one is in the subgroup
c) Only the second one is in the subgroup
For a) is the same to choose $5$ people out of $13$, i.e. ${13}choose{5}$
For b) is the same to choose $4$ people out of $13$ (because you have already one and you don't want the other), i.e. ${13}choose {4}$
For c) is the same as b)
Thus the solution is ${13}choose {5}$ +$2 $${13}choose{4}$ $=1287+2 cdot 715=1287+1430=2717$
answered Jan 20 at 11:13
user289143user289143
1,002313
$endgroup$
There are three possible types of subgroups:
a) None of the two persons are in the subgroup
b) Only the first one is in the subgroup
c) Only the second one is in the subgroup
For a) is the same to choose $5$ people out of $13$, i.e. ${13}choose{5}$
For b) is the same to choose $4$ people out of $13$ (because you have already one and you don't want the other), i.e. ${13}choose {4}$
For c) is the same as b)
Thus the solution is ${13}choose {5}$ +$2 $${13}choose{4}$ $=1287+2 cdot 715=1287+1430=2717$
answered Jan 20 at 11:13
user289143user289143
1,002313
answered Jan 20 at 11:13
user289143user289143
1,002313
answered Jan 20 at 11:13
user289143user289143
1,002313
answered Jan 20 at 11:13
user289143user289143
1,002313
1,002313
$begingroup$
Okay, so my mistake was that I calculated $15 choose 3$, I should have done $13 choose 3$ because I already have two people. Then I get to the same result as you did. So ${15 choose 5} - {13 choose 3}$ works out as well. Thank you very much.
$endgroup$
– Simon Schiller
Jan 20 at 11:21
add a comment |
$begingroup$
Okay, so my mistake was that I calculated $15 choose 3$, I should have done $13 choose 3$ because I already have two people. Then I get to the same result as you did. So ${15 choose 5} - {13 choose 3}$ works out as well. Thank you very much.
$endgroup$
– Simon Schiller
Jan 20 at 11:21
$begingroup$
Okay, so my mistake was that I calculated $15 choose 3$, I should have done $13 choose 3$ because I already have two people. Then I get to the same result as you did. So ${15 choose 5} - {13 choose 3}$ works out as well. Thank you very much.
$endgroup$
– Simon Schiller
Jan 20 at 11:21
$begingroup$
Okay, so my mistake was that I calculated $15 choose 3$, I should have done $13 choose 3$ because I already have two people. Then I get to the same result as you did. So ${15 choose 5} - {13 choose 3}$ works out as well. Thank you very much.
$endgroup$
– Simon Schiller
Jan 20 at 11:21
add a comment |
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$begingroup$
The negation of two people being in the subgroup together is that at least one of them is not in the subgroup, not that neither of them are in the subgroup
$endgroup$
– user289143
Jan 20 at 11:21