Let $k$ and $n$ be integers greater than 1. Then $(kn)!$ is not necessarily divisible by
- A. $(n!)^k$
- B. $(k!)^n$
- C. $n!\cdot k!$
- D. $2^{kn}$
I believe option D is correct and have a counter example for that.
Let $k=2 $ and $n=3$ then $(kn)!=6!=720$ is not divisible by $ 64=2^{2*3}$.
What I don't understand is that why options A, B, C necessarily divide $(kn)!$.
Thanks for help.
This can be seen from the fact that multinomial coefficients are integers : https://en.wikipedia.org/wiki/Multinomial_theorem#Multinomial_coefficients
Then $\frac{(nk)!}{(n!)^k} = \binom{nk}{n,n,\dots,n}$, $\frac{(nk)!}{(k!)^n} = \binom{nk}{k,k,\dots,k}$ and $\frac{(nk)!}{(n!)(k!)} = (nk-n-k)!\binom{nk}{n,k,nk-n-k}$.
For B. :$\binom {n+k} {k}=(n+k)!(n!!k!)^{-1}\in Z$ because it is the number of subsets of an $n+k$-element set that have exactly $k$ members each. So if $k>0$ then $k!$ divides the product of any $k$ consecutive positive integers, for if $k>0$ and $n\geq 0$ then $\binom {n+k} {k}= k!^{-1} \prod_{j=1}^k(n+j)).$ Therefore, for $n,i\geq 0 :$ $$A(i,k,n) = k!^{-1}\prod_{j=1}^k(i n+j)\in Z.$$ .So $k!^{-n}(nk)!=\prod_{i=0}^{n-1} A(i,k,n)\in Z.$
For A. : Interchange $k$ and $n$ in the argument for B.
For C. : $n +k\leq n k$ because $1\leq (n-1)(k-1)=n k -n-k+1.$ So $$(nk)!(n!k!)^{-1}=[(n k)!(n+k)!^{-1}] \binom {n+k} {k}\in Z.$$
$\frac{(kn)!}{n!^k}$ and $\frac{(kn)!}{k!^n}$ can both be recognized as multinomial coefficients which are integers.
Since $n,k>1$ also $\frac{(kn)!}{n!k!(kn-n-k)!}$ is a multinomial coefficient.
