1. Factorial division problem

As a last part of a solution of a second year algebra problem, I have to prove the identity:

a!b! | (a+b)!

a factorial times b factorial divides a+b factorial.
Anyone have any thoughts on how to tackle it? I tried 'writing out' the (a+b)! part, then cutting of either tail of a! or b!, but to no avail....

Tijmen

2. Hello, TiRune!

Prove that: .$\displaystyle a!b!\,|\,(a+b)!$
I don't have a proof, but I know intuitively that it's true.

If there are $\displaystyle a$ identical red balls and $\displaystyle b$ identical blue balls,
. . in how many ways can they be arranged in a row?

The answer is: .$\displaystyle {a+b\choose a,b} \:=\:\frac{(a+b)!}{a!\,b!}$ . which is an integer.

3. It suffices to show that every prime power which divides $\displaystyle a!b!$ also divides $\displaystyle (a+b)!$. Now you can use the fact that the highest power of $\displaystyle p$ which divides $\displaystyle n!$ is $\displaystyle p^{\sum_{j=1}^\infty\lfloor \frac{n}{p^j}\rfloor}$.

4. The product of n consecutive integers is divisible by n!

Consider the expansion of (a+b)!. It has (a+b) factors.

The product of the first a factors (or last a factors) is divisible by a!. The product of the last b factors (or first b factors) is divisible by b!.

Thus, (a+b)! is divisible by a!b!.

ie a!b! | (a+b)!

5. Originally Posted by TiRune
As a last part of a solution of a second year algebra problem, I have to prove the identity:

a!b! | (a+b)!

a factorial times b factorial divides a+b factorial.
Anyone have any thoughts on how to tackle it? I tried 'writing out' the (a+b)! part, then cutting of either tail of a! or b!, but to no avail....

Tijmen
$\displaystyle \frac{(a+b)!}{a!b!}=\frac{(a+b)!}{b!a!}=\frac{(a+b )!}{([a+b]-a)!a!}=\binom{a+b}{a}=n\ \in\ \mathbb{N}$

as this is "select 'a' elements from 'a+b' elements".