Proving factorial attribute

Question

Given that : $$ \sum_{i=1}^{k} a_i = n $$ I am asked to prove that: $$\prod_{j=1}^{k} a_j! $$ divises $n!$

I saw that it works for $k=1$, and for $k=2$ I tried :

$$\frac{n!}{a_1! a_2!} = \frac{(a_1 + a_2)!}{a_1! a_2!}$$

But came to nowhere.

How do I go on? I pretty sure it is by induction

Answer 1

HINT

Count the number of ways of arranging $n$ balls where $a_1$ balls are of color $1$, $a_2$ balls of color $2$ and in general, $a_j$ balls are of color $j$. (Note that $\displaystyle \sum_{j=1}^k a_j = n$.)

Answer 2

Hint for induction proof.

If $$\sum_{i=1}^{k} a_j = n$$ then

$$\sum_{i=1}^{k-1} b_i = n$$

where $b_1 = a_1 + a_2, b_2 = a_3, \dots$

Since OP does not seem interested in this hint, for sake of completeness...

$$\dfrac{n!}{\prod_{j=1}^{k} a_j!} = \dfrac{(a_1 + a_2)!}{a_1!a_2!}\cdot\dfrac{n!}{\prod_{j=1}^{k=1} b_j!}$$

which is the product of two integers.