Is $\frac{200!}{(10!)^{20} \cdot 19!}$ an integer or not? A friend of mine asked me to prove that $$\frac{200!}{(10!)^{20}}$$ is an integer
I used a basic example in which I assumed that there are $200$ objects places in $20$ boxes (which means that effectively there are $10$ objects in one box). One more condition that I adopted was that the boxes are distinguishable but the items within each box are not. Now the number of permutations possible for such an arrangement are : 
$$ \frac{200!}{\underbrace{10! \cdot 10! \cdot 10!\cdots 10!}_{\text{$20$ times}}}$$ 
$$\Rightarrow \frac{200!}{(10!) ^{20}}$$
Since these are just ways of arranging, we can be pretty sure that this number is an integer.
Then he made the problem more complex by adding a $19!$ in the denominator, thus making the problem:
Is $$\frac{200!}{(10!)^{20} \cdot 19!}$$ an integer or not?
The $19!$ in the denominator seemed to be pretty odd and hence I couldn't find any intuitive way to determine the thing. Can anybody please help me with the problem?
 A: You assumed the boxes were distinguishable, leading to $\frac{200!}{(10!)^{20}}$, ways to fill the boxes. If you make them indistinguishable, you merge the $20!$ ways of reordering the boxes into one, so that previous answer overcounts each way of filling indistinguishable boxes by a factor of $20!$. Therefore you are left with $\frac{200!}{(10!)^{20}}/20!$ ways to fill 20 indistinguishable boxes, which then must be an integer. After multiplying by $20$ it is of course still an integer.
A: We know that $\dfrac{(mn)!}{n!(m!)^n}$ is an integer for $m,n \in \Bbb N$ $^{(*)}$ . Let $n = 20$ and $m = 10$, then $\dfrac{(200)!}{20!(10!)^{20}}$ is an integer. 
Multiply by $20$, $\dfrac{(200)!}{19!(10!)^{20}}$ is an integer.
$(*)$ : Prove that $(mn)!$ is divisible by $(n!)\cdot(m!)^n$ 
A: Using induction, this answer says that
$$
\frac{(mn)!}{(m!)^nn!}=\prod_{k=1}^n\binom{mk-1}{m-1}
$$
Plug in $m=10$ and $n=20$ to get
$$
\frac{200!}{10!^{20}\,20!}=\prod_{k=1}^{20}\binom{10k-1}{9}
$$
Multiply by $20$ to get
$$
\frac{200!}{10!^{20}\,19!}=20\,\prod_{k=1}^{20}\binom{10k-1}{9}
$$

Another Approach
Note that
$$
\begin{align}
\binom{kn}{n}
&=\frac{(kn-n+1)(kn-n+2)\cdots(kn-1)\,kn}{1\cdot2\cdots(n-1)\,n}\\
&=\frac{(kn-n+1)(kn-n+2)\cdots(kn-1)\,k}{1\cdot2\cdots(n-1)}\\
&=\binom{kn-1}{n-1}\,k
\end{align}
$$
Therefore, since we can write a multinomial as a product of binomials,
$$
\begin{align}
\frac{(mn)!}{n!^m}
&=\prod_{k=1}^m\binom{kn}{n}\\
&=\prod_{k=1}^m\binom{kn-1}{n-1}\,k\\
&=m!\,\prod_{k=1}^m\binom{kn-1}{n-1}
\end{align}
$$
and so
$$
\frac{(mn)!}{n!^m\,m!}=\prod_{k=1}^m\binom{kn-1}{n-1}
$$
Plug in $m=20$ and $n=10$ and multiply by $20$ to get
$$
\frac{200!}{10!^{20}\,19!}=20\,\prod_{k=1}^{20}\binom{10k-1}{9}
$$
A: Consider $V=(10k+1)*....*(10k+9) $.
By your reasoning, ${10k+9 \choose 9}=(10k+1)*....*(10k+9)/9! $ is an integer.
And $10(k+1)/10(k+1)$ is an integer.  
So $(10k+1)*....*(10 (k+1)) $ is divisible by $9!*10*(k+1)=10!*(k+1)$.
So $200! $ is divisible by $10!*1*10!*2*10!*3*.....*10!*19=(10!)^{20}*19! $
A: Since I am pushing 70 yrs old, it seems appropriate to dinosaur-excerpt "Elementary Number Theory" 1938 (Uspensky and Heaslett).
For real # $r$, let $\lfloor r\rfloor \equiv$ the floor of $r$.
Let $p$ be any prime #.
Let $V_p(n) : n ~\in ~\mathbb{Z^+} ~\equiv~$
the largest exponent $\alpha$ such that $p^{\alpha} | n$. 
That is, if $\alpha = V_p(n),$ then $p^{(\alpha + 1)} \not | ~n.$
From Uspensky and Heaslett,
$V_p(n!) = \left\lfloor\frac{n}{p^1}\right\rfloor
~+~ \left\lfloor\frac{n}{p^2}\right\rfloor
~+~ \left\lfloor\frac{n}{p^3}\right\rfloor
~+~ \left\lfloor\frac{n}{p^4}\right\rfloor
\cdots $
Clearly, given two positive integers $A,B$,
$~\frac{A}{B}$ will be an integer
$\iff$ 
for every prime # $p$ that occurs in the prime factorization of
$B$, 
$V_p(B) \leq V_p(A).$
It is immediate, that given the OP's original question, the only prime #'s that need to be checked are those prime #'s that are $\leq 19.$  Further, you can see at a glance the prime #'s 11, 13, 17, and 19 can not pose a problem.
Therefore, the problem reduces to manually applying Uspenky and Heaslett's formula to the numerator and denominator of the OP's original query with respect to the prime #'s 2,3,5,7.
Empirically, they each check out okay.  Therefore, the fraction is an integer.
In your face $21^{\text{st}}$ century!
