Question about divisibility of the factorial I've found this question in the book devoted to Lie's groups. How to prove that $n!$ is devided by $n_{1}!n_{2}!...n_{k}!$, where $n=n_{1}+n_{2}+...+n_{k}$ and $n$, $n_{i}$ are natural numbers?
 A: There are several ways to do this.
One is to notice that
$$
\frac{n!}{n_{1}!n_{2}!...n_{k}!}
$$
is the multinomial coefficient, that is, the (integer) coefficient of $x_1^{n_1} \dots x_k^{n_k}$ in the expansion of
$$
(x_1 + \dots + x_k)^n.
$$
Another is to note that the symmetric group $S_n$, of order $n!$, contains a subgroup isomorphic to
$$
S_{n_1} \times \dots \times S_{n_k},
$$
of order 
$$
n_{1}!n_{2}!...n_{k}!
$$
The second argument is basically the same of @WouterZeldenthuis
A: Since $n!$ is the number of ways of ordering a set of size $n$, each $n_i!$ should be interpreted as the number of ways of ordering a subset of $n$. Simply apply a counting argument.
A: The combinatorial approaches are the "right" ones. However, it is a standard result that if $p$ is prime, then  the largest exponent $e$ such that $p^e$ divides $m!$ is given by
$$e=\left\lfloor \frac{m}{p}\right\rfloor+\left\lfloor \frac{m}{p^2}\right\rfloor+\left\lfloor \frac{m}{p^3}\right\rfloor+\left\lfloor \frac{m}{p^4}\right\rfloor+\cdots.$$
(Note that this is effectively a finite sum.) Now the desired divisibility result follows from the obvious fact that for any positive integer $b$, we have
$$\left\lfloor\frac{n_1}{b}\right\rfloor+ \left\lfloor\frac{n_2}{b}\right\rfloor+\cdots +\left\lfloor\frac{n_k}{b}\right\rfloor\le \left\lfloor \frac{n_1+n_2+\cdots+n_k}{b}\right\rfloor.$$
