Mixing
Number of ordered partitions of N into K distinct parts modulo P (with a “proof” using hyperplanes)
$begingroup$
I've come across a combinatorics problem where I'm fairly certain that a solution exists, yet I'm unable to find it.
I'm trying to find the number of vectors $(x_1,x_2,...,x_n)$ such that
$sum x_i = 0 mod P$ for some sufficiently large prime $P$ (i.e the resulting formula should hold for all but some finite number of $P$).
$x_i neq x_j, 1 leq i < j leq n$.
The answer should be $(p-1)(p-2)...(p-n+1)$. This is because it can be found implicitly from a problem in my textbook.
Observe the hyperplane arrangement given by $x_i = x_j, 1 leq i < j leq n, sum^n_{i=0} x_i = 0$. Let $h_0 = (sum x_i = 0)$ (the plane). If $A' = A - h_0$, $A'' = { H cap h_0, H in A'}$ then $X_{A}(x) = X_{A'}(x) - X_{A''}(x)$ where $X_A(x)$ is the characteristic polynomial for $A$ etc.
The problem in question is to find $X_A$,and the answer is $(p-1)*(p-1)(p-2)...(p-n+1)$. $X_{A'}$ is easy to find, as $p(p-1)...(p-n+1)$. Thus $X_{A''}=(p-1)(p-2)...(p-n+1)$. Due to the finite field method, finding $X_{A''}$ is equivalent to finding the mentioned partitions.
What I would like to do is find a combinatorial proof of the number of partitions. However, I'm stumped.
combinatorics discrete-mathematics order-theory integer-partitions
asked Jan 2 at 15:46
Sven SvenssonSven Svensson
132
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add a comment |
$begingroup$
I've come across a combinatorics problem where I'm fairly certain that a solution exists, yet I'm unable to find it.
I'm trying to find the number of vectors $(x_1,x_2,...,x_n)$ such that
$sum x_i = 0 mod P$ for some sufficiently large prime $P$ (i.e the resulting formula should hold for all but some finite number of $P$).
$x_i neq x_j, 1 leq i < j leq n$.
The answer should be $(p-1)(p-2)...(p-n+1)$. This is because it can be found implicitly from a problem in my textbook.
Observe the hyperplane arrangement given by $x_i = x_j, 1 leq i < j leq n, sum^n_{i=0} x_i = 0$. Let $h_0 = (sum x_i = 0)$ (the plane). If $A' = A - h_0$, $A'' = { H cap h_0, H in A'}$ then $X_{A}(x) = X_{A'}(x) - X_{A''}(x)$ where $X_A(x)$ is the characteristic polynomial for $A$ etc.
The problem in question is to find $X_A$,and the answer is $(p-1)*(p-1)(p-2)...(p-n+1)$. $X_{A'}$ is easy to find, as $p(p-1)...(p-n+1)$. Thus $X_{A''}=(p-1)(p-2)...(p-n+1)$. Due to the finite field method, finding $X_{A''}$ is equivalent to finding the mentioned partitions.
What I would like to do is find a combinatorial proof of the number of partitions. However, I'm stumped.
combinatorics discrete-mathematics order-theory integer-partitions
asked Jan 2 at 15:46
Sven SvenssonSven Svensson
132
$endgroup$
add a comment |
$begingroup$
I've come across a combinatorics problem where I'm fairly certain that a solution exists, yet I'm unable to find it.
I'm trying to find the number of vectors $(x_1,x_2,...,x_n)$ such that
$sum x_i = 0 mod P$ for some sufficiently large prime $P$ (i.e the resulting formula should hold for all but some finite number of $P$).
$x_i neq x_j, 1 leq i < j leq n$.
The answer should be $(p-1)(p-2)...(p-n+1)$. This is because it can be found implicitly from a problem in my textbook.
Observe the hyperplane arrangement given by $x_i = x_j, 1 leq i < j leq n, sum^n_{i=0} x_i = 0$. Let $h_0 = (sum x_i = 0)$ (the plane). If $A' = A - h_0$, $A'' = { H cap h_0, H in A'}$ then $X_{A}(x) = X_{A'}(x) - X_{A''}(x)$ where $X_A(x)$ is the characteristic polynomial for $A$ etc.
The problem in question is to find $X_A$,and the answer is $(p-1)*(p-1)(p-2)...(p-n+1)$. $X_{A'}$ is easy to find, as $p(p-1)...(p-n+1)$. Thus $X_{A''}=(p-1)(p-2)...(p-n+1)$. Due to the finite field method, finding $X_{A''}$ is equivalent to finding the mentioned partitions.
What I would like to do is find a combinatorial proof of the number of partitions. However, I'm stumped.
combinatorics discrete-mathematics order-theory integer-partitions
asked Jan 2 at 15:46
Sven SvenssonSven Svensson
132
$endgroup$
I've come across a combinatorics problem where I'm fairly certain that a solution exists, yet I'm unable to find it.
I'm trying to find the number of vectors $(x_1,x_2,...,x_n)$ such that
$sum x_i = 0 mod P$ for some sufficiently large prime $P$ (i.e the resulting formula should hold for all but some finite number of $P$).
$x_i neq x_j, 1 leq i < j leq n$.
The answer should be $(p-1)(p-2)...(p-n+1)$. This is because it can be found implicitly from a problem in my textbook.
Observe the hyperplane arrangement given by $x_i = x_j, 1 leq i < j leq n, sum^n_{i=0} x_i = 0$. Let $h_0 = (sum x_i = 0)$ (the plane). If $A' = A - h_0$, $A'' = { H cap h_0, H in A'}$ then $X_{A}(x) = X_{A'}(x) - X_{A''}(x)$ where $X_A(x)$ is the characteristic polynomial for $A$ etc.
The problem in question is to find $X_A$,and the answer is $(p-1)*(p-1)(p-2)...(p-n+1)$. $X_{A'}$ is easy to find, as $p(p-1)...(p-n+1)$. Thus $X_{A''}=(p-1)(p-2)...(p-n+1)$. Due to the finite field method, finding $X_{A''}$ is equivalent to finding the mentioned partitions.
What I would like to do is find a combinatorial proof of the number of partitions. However, I'm stumped.
combinatorics discrete-mathematics order-theory integer-partitions
combinatorics discrete-mathematics order-theory integer-partitions
asked Jan 2 at 15:46
Sven SvenssonSven Svensson
132
asked Jan 2 at 15:46
Sven SvenssonSven Svensson
132
asked Jan 2 at 15:46
Sven SvenssonSven Svensson
132
asked Jan 2 at 15:46
Sven SvenssonSven Svensson
132
asked Jan 2 at 15:46
Sven SvenssonSven Svensson
132
132
add a comment |
add a comment |
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Here is a simple combinatorial proof that does not use hyperplanes.
The number of vectors in $mathbb F_p^n$ whose entries are pairwise different is $p(p-1)(p-2)cdots(p-n+1)$. Given such a vector, $defx{{bf x}}x$, let $f(x)$ be the vector obtained by adding one to each entry. As long as $n$ and $p$ are corprime, which occurs whenever $p>n$, the sums of the entries of the vectors $$x,f(x),f^2(x),dots,f^{p-1}(x)$$ will each have a different remainder modulo $p$. Therefore, exactly one of them will have a sum whose remainder is zero. Since the set of all vectors with pairwise different entries are partitioned into groups like this, the number of such vectors whose sum congruent to zero is $1/p$ times the total number, which is $(p-1)(p-2)dots(p-n+1)$.
answered Jan 2 at 18:18
Mike EarnestMike Earnest
21k11951
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$begingroup$
Here is a simple combinatorial proof that does not use hyperplanes.
The number of vectors in $mathbb F_p^n$ whose entries are pairwise different is $p(p-1)(p-2)cdots(p-n+1)$. Given such a vector, $defx{{bf x}}x$, let $f(x)$ be the vector obtained by adding one to each entry. As long as $n$ and $p$ are corprime, which occurs whenever $p>n$, the sums of the entries of the vectors $$x,f(x),f^2(x),dots,f^{p-1}(x)$$ will each have a different remainder modulo $p$. Therefore, exactly one of them will have a sum whose remainder is zero. Since the set of all vectors with pairwise different entries are partitioned into groups like this, the number of such vectors whose sum congruent to zero is $1/p$ times the total number, which is $(p-1)(p-2)dots(p-n+1)$.
answered Jan 2 at 18:18
Mike EarnestMike Earnest
21k11951
$endgroup$
add a comment |
$begingroup$
Here is a simple combinatorial proof that does not use hyperplanes.
The number of vectors in $mathbb F_p^n$ whose entries are pairwise different is $p(p-1)(p-2)cdots(p-n+1)$. Given such a vector, $defx{{bf x}}x$, let $f(x)$ be the vector obtained by adding one to each entry. As long as $n$ and $p$ are corprime, which occurs whenever $p>n$, the sums of the entries of the vectors $$x,f(x),f^2(x),dots,f^{p-1}(x)$$ will each have a different remainder modulo $p$. Therefore, exactly one of them will have a sum whose remainder is zero. Since the set of all vectors with pairwise different entries are partitioned into groups like this, the number of such vectors whose sum congruent to zero is $1/p$ times the total number, which is $(p-1)(p-2)dots(p-n+1)$.
answered Jan 2 at 18:18
Mike EarnestMike Earnest
21k11951
$endgroup$
add a comment |
$begingroup$
Here is a simple combinatorial proof that does not use hyperplanes.
The number of vectors in $mathbb F_p^n$ whose entries are pairwise different is $p(p-1)(p-2)cdots(p-n+1)$. Given such a vector, $defx{{bf x}}x$, let $f(x)$ be the vector obtained by adding one to each entry. As long as $n$ and $p$ are corprime, which occurs whenever $p>n$, the sums of the entries of the vectors $$x,f(x),f^2(x),dots,f^{p-1}(x)$$ will each have a different remainder modulo $p$. Therefore, exactly one of them will have a sum whose remainder is zero. Since the set of all vectors with pairwise different entries are partitioned into groups like this, the number of such vectors whose sum congruent to zero is $1/p$ times the total number, which is $(p-1)(p-2)dots(p-n+1)$.
answered Jan 2 at 18:18
Mike EarnestMike Earnest
21k11951
$endgroup$
Here is a simple combinatorial proof that does not use hyperplanes.
The number of vectors in $mathbb F_p^n$ whose entries are pairwise different is $p(p-1)(p-2)cdots(p-n+1)$. Given such a vector, $defx{{bf x}}x$, let $f(x)$ be the vector obtained by adding one to each entry. As long as $n$ and $p$ are corprime, which occurs whenever $p>n$, the sums of the entries of the vectors $$x,f(x),f^2(x),dots,f^{p-1}(x)$$ will each have a different remainder modulo $p$. Therefore, exactly one of them will have a sum whose remainder is zero. Since the set of all vectors with pairwise different entries are partitioned into groups like this, the number of such vectors whose sum congruent to zero is $1/p$ times the total number, which is $(p-1)(p-2)dots(p-n+1)$.
answered Jan 2 at 18:18
Mike EarnestMike Earnest
21k11951
answered Jan 2 at 18:18
Mike EarnestMike Earnest
21k11951
answered Jan 2 at 18:18
Mike EarnestMike Earnest
21k11951
answered Jan 2 at 18:18
Mike EarnestMike Earnest
21k11951
21k11951
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