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How do you proceed if you are required to prove for any natural number $n$ that $$\frac{n^2!}{(n!)^n}$$ is an integer. Here the ! sign represents factorial. I got absolutely no leads on this problem. Any hint or help would be appreciated.

najayaz
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    Have you tried using the fundamental theorem of arithmetic? It suffices to show that for each prime the power occurring in $n^2!$ is higher than the one in $(n!)^n$. – Joonas Ilmavirta Oct 03 '14 at 09:45
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    Note that some of your Readers overlooked or ignored the request mentioned only in the title for proof by combinatorial methods. It a good idea to include such requirements in the body of the Question as well. – hardmath Oct 03 '14 at 14:21
  • @JoonasIlmavirta: I tried to come up with a proof based on your suggestion, even though it wouldn't be a good answer (not combinatorics). It turned out to be harder than I expected to put a bound on the highest power of a prime factor! I ended up with an unproven conjecture that the product of any $x$ consecutive counting numbers is divisible by each of ${1,\ldots,x}$ (or by $x!$, equivalently) and got stuck there... – Tim Pederick Oct 03 '14 at 15:42
  • I had indeed overlooked the request to use combinatorial methods because it was hidden in the title. @TimPederick, the arithmetic proof indeed seems to require something like your conjecture: Take any prime $p$. If its multiples occur $k$ times in ${1,2,\dots,n}$ (counting squares as two occurrences etc.), then they also occur at least $k$ times in ${a+1,\dots,a+n}$ for any $a\geq0$. Therefore $(a+1)\cdots(a+n)/n!$ is an integer for all $a\geq0$. Let $a$ range in ${0,\dots,n-1}$ and take the product of these integers. – Joonas Ilmavirta Oct 03 '14 at 17:03
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    Possible duplicate of prove that $\frac{(2n)!}{(n!)^2}$ is even if $n$ is a positive integer – Davood Sep 04 '17 at 13:36
  • why are you flagging 3 year old questions with accepted answers, though? – najayaz Sep 05 '17 at 00:38

4 Answers4

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How many ways are there to order a "word" made with $n$ letters, each of which occur $n$ times?

Platehead
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We'll show that the above number is the size of a set with integer number of elements.

Assume we have $n$ types of balls, and from each type we have $n$ balls. Obviously, there are $n^2$ balls. Let's set them up in a line; obviously, there is an integer number of ways to do that.

First we set them all in a line as they are different- permutation over $n^2$ different elements is defined as $n^2!$, now divide by number of substitutions for each type - $n!$ for each type we get just what we wanted: $$\frac{n^2!}{(n!)^n} $$

wchargin
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Snufsan
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The product of $n$ consecutive integers is divisible by $n!$: $$ (m+1)(m+2)\cdots(m+n)=\binom{m+n}{n}n! $$

$n^2!$ is a product of $n$ products of $n$ consecutive integers: $$ [1 \cdot 2 \cdots n]\dot[(n+1)(n+2)\cdots(n+n)]]\cdots $$

Explicily, $$ n^2! = \binom{n}{n}\binom{2n}{n}\cdots \binom{n^2}{n}(n!)^n $$

lhf
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a useful principle is that for any natural numbers $m,n$ we have: $$ F_{n,0} | F_{n,m} $$ where $$ F(n,m) = \prod_{j=1+m}^{n+m} j $$ with this notation we have $F(n,0) = n!$ and $$ (kn)! = \prod_{j=0}^{k-1} F(n,kn) $$ setting $k=n$ gives the required result

David Holden
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