Calculating the number of permutations that also contain a specific subset
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If I have a set of {1,2,3,4,5,6} and I need to calculate the number of permutations where I take 3 numbers then I know that the formula for the total number of permutations is:
$frac{n!}{(n-r)!} = frac{6!}{(6-3)!} = 120 $
How can I calculate the number of permutations that contain 1 and 2?
Eg. {1,2,3}, {1,3,2}, {2,1,4} etc.
combinatorics permutations
asked Jan 14 at 10:16
KevinKevin
31
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add a comment |
$begingroup$
If I have a set of {1,2,3,4,5,6} and I need to calculate the number of permutations where I take 3 numbers then I know that the formula for the total number of permutations is:
$frac{n!}{(n-r)!} = frac{6!}{(6-3)!} = 120 $
How can I calculate the number of permutations that contain 1 and 2?
Eg. {1,2,3}, {1,3,2}, {2,1,4} etc.
combinatorics permutations
asked Jan 14 at 10:16
KevinKevin
31
$endgroup$
add a comment |
$begingroup$
If I have a set of {1,2,3,4,5,6} and I need to calculate the number of permutations where I take 3 numbers then I know that the formula for the total number of permutations is:
$frac{n!}{(n-r)!} = frac{6!}{(6-3)!} = 120 $
How can I calculate the number of permutations that contain 1 and 2?
Eg. {1,2,3}, {1,3,2}, {2,1,4} etc.
combinatorics permutations
asked Jan 14 at 10:16
KevinKevin
31
$endgroup$
If I have a set of {1,2,3,4,5,6} and I need to calculate the number of permutations where I take 3 numbers then I know that the formula for the total number of permutations is:
$frac{n!}{(n-r)!} = frac{6!}{(6-3)!} = 120 $
How can I calculate the number of permutations that contain 1 and 2?
Eg. {1,2,3}, {1,3,2}, {2,1,4} etc.
combinatorics permutations
combinatorics permutations
asked Jan 14 at 10:16
KevinKevin
31
asked Jan 14 at 10:16
KevinKevin
31
asked Jan 14 at 10:16
KevinKevin
31
asked Jan 14 at 10:16
KevinKevin
31
asked Jan 14 at 10:16
KevinKevin
31
31
add a comment |
add a comment |
1 Answer
1
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$begingroup$
You calculate the number $S$ of different sets, including your subset. So to make set of size 3, if we already know 2 elements, we need to choose 1 from the last 4 elements:
$$
S = {6-2 choose 3-2}={4choose 1}=4
$$
Then for each of those sets you calculate all the permutations of its elements, it's $3!$
So the number of permutations of size $r=3$ from set of size $n=6$ which contatin given $k=2$ elements is:
$$
N={n-kchoose r-k}r!=frac{(n-k)! r!}{(n-r)!(r-k!)}
$$
answered Jan 14 at 10:38
Vasily MitchVasily Mitch
2,6891312
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add a comment |
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1 Answer
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$begingroup$
You calculate the number $S$ of different sets, including your subset. So to make set of size 3, if we already know 2 elements, we need to choose 1 from the last 4 elements:
$$
S = {6-2 choose 3-2}={4choose 1}=4
$$
Then for each of those sets you calculate all the permutations of its elements, it's $3!$
So the number of permutations of size $r=3$ from set of size $n=6$ which contatin given $k=2$ elements is:
$$
N={n-kchoose r-k}r!=frac{(n-k)! r!}{(n-r)!(r-k!)}
$$
answered Jan 14 at 10:38
Vasily MitchVasily Mitch
2,6891312
$endgroup$
add a comment |
$begingroup$
You calculate the number $S$ of different sets, including your subset. So to make set of size 3, if we already know 2 elements, we need to choose 1 from the last 4 elements:
$$
S = {6-2 choose 3-2}={4choose 1}=4
$$
Then for each of those sets you calculate all the permutations of its elements, it's $3!$
So the number of permutations of size $r=3$ from set of size $n=6$ which contatin given $k=2$ elements is:
$$
N={n-kchoose r-k}r!=frac{(n-k)! r!}{(n-r)!(r-k!)}
$$
answered Jan 14 at 10:38
Vasily MitchVasily Mitch
2,6891312
$endgroup$
add a comment |
$begingroup$
You calculate the number $S$ of different sets, including your subset. So to make set of size 3, if we already know 2 elements, we need to choose 1 from the last 4 elements:
$$
S = {6-2 choose 3-2}={4choose 1}=4
$$
Then for each of those sets you calculate all the permutations of its elements, it's $3!$
So the number of permutations of size $r=3$ from set of size $n=6$ which contatin given $k=2$ elements is:
$$
N={n-kchoose r-k}r!=frac{(n-k)! r!}{(n-r)!(r-k!)}
$$
answered Jan 14 at 10:38
Vasily MitchVasily Mitch
2,6891312
$endgroup$
You calculate the number $S$ of different sets, including your subset. So to make set of size 3, if we already know 2 elements, we need to choose 1 from the last 4 elements:
$$
S = {6-2 choose 3-2}={4choose 1}=4
$$
Then for each of those sets you calculate all the permutations of its elements, it's $3!$
So the number of permutations of size $r=3$ from set of size $n=6$ which contatin given $k=2$ elements is:
$$
N={n-kchoose r-k}r!=frac{(n-k)! r!}{(n-r)!(r-k!)}
$$
answered Jan 14 at 10:38
Vasily MitchVasily Mitch
2,6891312
answered Jan 14 at 10:38
Vasily MitchVasily Mitch
2,6891312
answered Jan 14 at 10:38
Vasily MitchVasily Mitch
2,6891312
answered Jan 14 at 10:38
Vasily MitchVasily Mitch
2,6891312
2,6891312
add a comment |
add a comment |
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