Find number of permutations with 2 inversions
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Suppose $sigma=(a_1,a_2,a_3...,a_n)$ be a permutation of $(1,2,3,...n)$. A pair $(a_i,a_j)$ is said to correspond to an inversion of $sigma$, if $i<j$ but $a_i > a_j$. How many permutations of $(1,2,3...n)$, $(ngeq 3)$, have exactly $2$ inversions?
Example: In the permutation $(2,4,5,3,1)$, there are 6 inversions corresponding to the pairs $(2,1),(4,3), (4,1), (5,3), (5,1), (3,1)$.
Can anyone give a recursive proof for this one? I tried to get one, but I dont get a proper solution..
combinatorics
asked Jan 14 at 16:52
YellowYellow
16011
$endgroup$
|
show 1 more comment
$begingroup$
Suppose $sigma=(a_1,a_2,a_3...,a_n)$ be a permutation of $(1,2,3,...n)$. A pair $(a_i,a_j)$ is said to correspond to an inversion of $sigma$, if $i<j$ but $a_i > a_j$. How many permutations of $(1,2,3...n)$, $(ngeq 3)$, have exactly $2$ inversions?
Example: In the permutation $(2,4,5,3,1)$, there are 6 inversions corresponding to the pairs $(2,1),(4,3), (4,1), (5,3), (5,1), (3,1)$.
Can anyone give a recursive proof for this one? I tried to get one, but I dont get a proper solution..
combinatorics
asked Jan 14 at 16:52
YellowYellow
16011
$endgroup$
$begingroup$
What is the answer for $n=3?$ For $n=4?$
$endgroup$
– saulspatz
Jan 14 at 16:54
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Does it have to be recursive? I think a direct proof is easier
$endgroup$
– Ross Millikan
Jan 14 at 17:06
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I thought recursive proofs were easier, but if there is an easier proof than a recursive one, I am still OK with it.
$endgroup$
– Yellow
Jan 14 at 17:15
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They can be, but not always. Have you done the $n=3$ and $n=4$ cases like saulspatz suggested?
$endgroup$
– Ross Millikan
Jan 14 at 17:20
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Nope,( as I said earlier, I was trying to get a recursive one without knowing there were other easier proofs, so I left that computation part halfway...) I am doing it.
$endgroup$
– Yellow
Jan 14 at 17:26
|
show 1 more comment
$begingroup$
Suppose $sigma=(a_1,a_2,a_3...,a_n)$ be a permutation of $(1,2,3,...n)$. A pair $(a_i,a_j)$ is said to correspond to an inversion of $sigma$, if $i<j$ but $a_i > a_j$. How many permutations of $(1,2,3...n)$, $(ngeq 3)$, have exactly $2$ inversions?
Example: In the permutation $(2,4,5,3,1)$, there are 6 inversions corresponding to the pairs $(2,1),(4,3), (4,1), (5,3), (5,1), (3,1)$.
Can anyone give a recursive proof for this one? I tried to get one, but I dont get a proper solution..
combinatorics
asked Jan 14 at 16:52
YellowYellow
16011
$endgroup$
Suppose $sigma=(a_1,a_2,a_3...,a_n)$ be a permutation of $(1,2,3,...n)$. A pair $(a_i,a_j)$ is said to correspond to an inversion of $sigma$, if $i<j$ but $a_i > a_j$. How many permutations of $(1,2,3...n)$, $(ngeq 3)$, have exactly $2$ inversions?
Example: In the permutation $(2,4,5,3,1)$, there are 6 inversions corresponding to the pairs $(2,1),(4,3), (4,1), (5,3), (5,1), (3,1)$.
Can anyone give a recursive proof for this one? I tried to get one, but I dont get a proper solution..
combinatorics
combinatorics
asked Jan 14 at 16:52
YellowYellow
16011
asked Jan 14 at 16:52
YellowYellow
16011
asked Jan 14 at 16:52
YellowYellow
16011
asked Jan 14 at 16:52
YellowYellow
16011
asked Jan 14 at 16:52
YellowYellow
16011
16011
$begingroup$
What is the answer for $n=3?$ For $n=4?$
$endgroup$
– saulspatz
Jan 14 at 16:54
$begingroup$
Does it have to be recursive? I think a direct proof is easier
$endgroup$
– Ross Millikan
Jan 14 at 17:06
$begingroup$
I thought recursive proofs were easier, but if there is an easier proof than a recursive one, I am still OK with it.
$endgroup$
– Yellow
Jan 14 at 17:15
$begingroup$
They can be, but not always. Have you done the $n=3$ and $n=4$ cases like saulspatz suggested?
$endgroup$
– Ross Millikan
Jan 14 at 17:20
$begingroup$
Nope,( as I said earlier, I was trying to get a recursive one without knowing there were other easier proofs, so I left that computation part halfway...) I am doing it.
$endgroup$
– Yellow
Jan 14 at 17:26
|
show 1 more comment
$begingroup$
What is the answer for $n=3?$ For $n=4?$
$endgroup$
– saulspatz
Jan 14 at 16:54
$begingroup$
Does it have to be recursive? I think a direct proof is easier
$endgroup$
– Ross Millikan
Jan 14 at 17:06
$begingroup$
I thought recursive proofs were easier, but if there is an easier proof than a recursive one, I am still OK with it.
$endgroup$
– Yellow
Jan 14 at 17:15
$begingroup$
They can be, but not always. Have you done the $n=3$ and $n=4$ cases like saulspatz suggested?
$endgroup$
– Ross Millikan
Jan 14 at 17:20
$begingroup$
Nope,( as I said earlier, I was trying to get a recursive one without knowing there were other easier proofs, so I left that computation part halfway...) I am doing it.
$endgroup$
– Yellow
Jan 14 at 17:26
$begingroup$
What is the answer for $n=3?$ For $n=4?$
$endgroup$
– saulspatz
Jan 14 at 16:54
$begingroup$
What is the answer for $n=3?$ For $n=4?$
$endgroup$
– saulspatz
Jan 14 at 16:54
$begingroup$
Does it have to be recursive? I think a direct proof is easier
$endgroup$
– Ross Millikan
Jan 14 at 17:06
$begingroup$
Does it have to be recursive? I think a direct proof is easier
$endgroup$
– Ross Millikan
Jan 14 at 17:06
$begingroup$
I thought recursive proofs were easier, but if there is an easier proof than a recursive one, I am still OK with it.
$endgroup$
– Yellow
Jan 14 at 17:15
$begingroup$
I thought recursive proofs were easier, but if there is an easier proof than a recursive one, I am still OK with it.
$endgroup$
– Yellow
Jan 14 at 17:15
$begingroup$
They can be, but not always. Have you done the $n=3$ and $n=4$ cases like saulspatz suggested?
$endgroup$
– Ross Millikan
Jan 14 at 17:20
$begingroup$
They can be, but not always. Have you done the $n=3$ and $n=4$ cases like saulspatz suggested?
$endgroup$
– Ross Millikan
Jan 14 at 17:20
$begingroup$
Nope,( as I said earlier, I was trying to get a recursive one without knowing there were other easier proofs, so I left that computation part halfway...) I am doing it.
$endgroup$
– Yellow
Jan 14 at 17:26
$begingroup$
Nope,( as I said earlier, I was trying to get a recursive one without knowing there were other easier proofs, so I left that computation part halfway...) I am doing it.
$endgroup$
– Yellow
Jan 14 at 17:26
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Consider the following method of building a permutation. For each number in ${1,2,dots,n}$ in increasing order, insert it anywhere in the line of the previously made numbers. The $i^{th}$ number can be inserted in $i$ places, and the number of inversions that insertion will add is anywhere between $0$ and $i-1$. For example, there is only one place to put $1$, effectively creating the list. There are $2$ places to put $2$, either before or after the $1$.
How many ways are there to end up with two inversions? There are two ways this can happen; either there were two steps which each created one inversion (and every other step made zero), or there was only one step which created two inversions. In the former case, there are $binom{n-1}2$ ways to choose the two steps which created the inversions (as the first step, where $1$ is inserted into an empty list, cannot create any inversions), and there is are $n-2$ ways to choose a step where two inversions are created (as this cannot happen on the first or second steps). Therefore, the total number of ways is
$$
binom{n-1}2+n-2 = frac{(n+1)(n-2)}2
$$
answered Jan 14 at 19:06
Mike EarnestMike Earnest
28.2k22152
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add a comment |
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$begingroup$
Consider the following method of building a permutation. For each number in ${1,2,dots,n}$ in increasing order, insert it anywhere in the line of the previously made numbers. The $i^{th}$ number can be inserted in $i$ places, and the number of inversions that insertion will add is anywhere between $0$ and $i-1$. For example, there is only one place to put $1$, effectively creating the list. There are $2$ places to put $2$, either before or after the $1$.
How many ways are there to end up with two inversions? There are two ways this can happen; either there were two steps which each created one inversion (and every other step made zero), or there was only one step which created two inversions. In the former case, there are $binom{n-1}2$ ways to choose the two steps which created the inversions (as the first step, where $1$ is inserted into an empty list, cannot create any inversions), and there is are $n-2$ ways to choose a step where two inversions are created (as this cannot happen on the first or second steps). Therefore, the total number of ways is
$$
binom{n-1}2+n-2 = frac{(n+1)(n-2)}2
$$
answered Jan 14 at 19:06
Mike EarnestMike Earnest
28.2k22152
$endgroup$
add a comment |
$begingroup$
Consider the following method of building a permutation. For each number in ${1,2,dots,n}$ in increasing order, insert it anywhere in the line of the previously made numbers. The $i^{th}$ number can be inserted in $i$ places, and the number of inversions that insertion will add is anywhere between $0$ and $i-1$. For example, there is only one place to put $1$, effectively creating the list. There are $2$ places to put $2$, either before or after the $1$.
How many ways are there to end up with two inversions? There are two ways this can happen; either there were two steps which each created one inversion (and every other step made zero), or there was only one step which created two inversions. In the former case, there are $binom{n-1}2$ ways to choose the two steps which created the inversions (as the first step, where $1$ is inserted into an empty list, cannot create any inversions), and there is are $n-2$ ways to choose a step where two inversions are created (as this cannot happen on the first or second steps). Therefore, the total number of ways is
$$
binom{n-1}2+n-2 = frac{(n+1)(n-2)}2
$$
answered Jan 14 at 19:06
Mike EarnestMike Earnest
28.2k22152
$endgroup$
add a comment |
$begingroup$
Consider the following method of building a permutation. For each number in ${1,2,dots,n}$ in increasing order, insert it anywhere in the line of the previously made numbers. The $i^{th}$ number can be inserted in $i$ places, and the number of inversions that insertion will add is anywhere between $0$ and $i-1$. For example, there is only one place to put $1$, effectively creating the list. There are $2$ places to put $2$, either before or after the $1$.
How many ways are there to end up with two inversions? There are two ways this can happen; either there were two steps which each created one inversion (and every other step made zero), or there was only one step which created two inversions. In the former case, there are $binom{n-1}2$ ways to choose the two steps which created the inversions (as the first step, where $1$ is inserted into an empty list, cannot create any inversions), and there is are $n-2$ ways to choose a step where two inversions are created (as this cannot happen on the first or second steps). Therefore, the total number of ways is
$$
binom{n-1}2+n-2 = frac{(n+1)(n-2)}2
$$
answered Jan 14 at 19:06
Mike EarnestMike Earnest
28.2k22152
$endgroup$
Consider the following method of building a permutation. For each number in ${1,2,dots,n}$ in increasing order, insert it anywhere in the line of the previously made numbers. The $i^{th}$ number can be inserted in $i$ places, and the number of inversions that insertion will add is anywhere between $0$ and $i-1$. For example, there is only one place to put $1$, effectively creating the list. There are $2$ places to put $2$, either before or after the $1$.
How many ways are there to end up with two inversions? There are two ways this can happen; either there were two steps which each created one inversion (and every other step made zero), or there was only one step which created two inversions. In the former case, there are $binom{n-1}2$ ways to choose the two steps which created the inversions (as the first step, where $1$ is inserted into an empty list, cannot create any inversions), and there is are $n-2$ ways to choose a step where two inversions are created (as this cannot happen on the first or second steps). Therefore, the total number of ways is
$$
binom{n-1}2+n-2 = frac{(n+1)(n-2)}2
$$
answered Jan 14 at 19:06
Mike EarnestMike Earnest
28.2k22152
answered Jan 14 at 19:06
Mike EarnestMike Earnest
28.2k22152
answered Jan 14 at 19:06
Mike EarnestMike Earnest
28.2k22152
answered Jan 14 at 19:06
Mike EarnestMike Earnest
28.2k22152
28.2k22152
add a comment |
add a comment |
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$begingroup$
What is the answer for $n=3?$ For $n=4?$
$endgroup$
– saulspatz
Jan 14 at 16:54
$begingroup$
Does it have to be recursive? I think a direct proof is easier
$endgroup$
– Ross Millikan
Jan 14 at 17:06
$begingroup$
I thought recursive proofs were easier, but if there is an easier proof than a recursive one, I am still OK with it.
$endgroup$
– Yellow
Jan 14 at 17:15
$begingroup$
They can be, but not always. Have you done the $n=3$ and $n=4$ cases like saulspatz suggested?
$endgroup$
– Ross Millikan
Jan 14 at 17:20
$begingroup$
Nope,( as I said earlier, I was trying to get a recursive one without knowing there were other easier proofs, so I left that computation part halfway...) I am doing it.
$endgroup$
– Yellow
Jan 14 at 17:26