Looking for a non-combinatorial proof that $a! \cdot b! \mid (a+b)!$

Question

(I use $a$ and $b$ to denote natural numbers.)

Question. Without appealing to the combinatorial interpretation of $$\frac{(a+b)!}{a! b!}$$ as a multinomial coefficients, is there a proof that for all $a$ and $b$, we have $$a! \cdot b! \mid (a+b)! \qquad?$$

Basically, I want a proof that just uses some clever algebra.

I was thinking that maybe we can use modular arithmetic, and try to understand the value of $(a+b)!$ modulo $a! \cdot b!$, and eventually show that this is $0$.

Ideas, anyone?

Answer 1

You can use $\nu_{p}(n!)=\sum \limits_{k\ge 1}\left[\dfrac{n}{p^k}\right]$ and that $[a+b]\ge [a]+[b]$

Answer 2

If we use as a lemma that the product of $k$ consecutive integers is divisible by $k!$ (proven e.g. here: The product of n consecutive integers is divisible by n factorial), we see that $(a+b)!/a!$ factors into the $b$ consecutive integers $a+1,\ldots,a+b$, hence is divisible by $b!$. Then $a!b!|(a+b)!$.

Answer 3

Possibly the answer that you don't want:

$$\frac{(a+0)!}{a!\cdot 0!}=1=\frac{(0+b)!}{0!\cdot b!}$$

and

$$\frac{(a+b)!}{a!\cdot b!}=\frac{(a+b)(a+b-1)!}{a!\cdot b!}=\frac{(a-1+b)!}{(a-1)!\cdot b!}+\frac{(a+b-1)!}{a!\cdot (b-1)!}.$$

Then by induction, Pascal's triangle is made of integers.

Answer 4

With the settings

\begin{align*} [k]_q:=\frac{1-q^k}{1-q}\qquad\text{and}\qquad [k]_q!:=\prod_{j=1}^{k}[j]_q=\prod_{j=1}^{k}\frac{1-q^j}{1-q} \end{align*}

the q-binomial coefficient $\begin{bmatrix}a+b\\a\end{bmatrix}_q$ is defined as

\begin{align*} \begin{bmatrix}a+b\\a\end{bmatrix}_q:=\frac{[a+b]_q!}{[a]_q![b]_q!} \end{align*}

Since the following is valid \begin{align*} \begin{bmatrix}a+b\\a\end{bmatrix}_q&=\begin{bmatrix}a+b-1\\a\end{bmatrix}_q+q^{b}\begin{bmatrix}a+b-1\\a-1\end{bmatrix}_q\\ \begin{bmatrix}a\\0\end{bmatrix}_q&=\begin{bmatrix}a\\a\end{bmatrix}_q=1 \end{align*} the $q$-binomial coefficients are polynomials in $q$ with non-negative integer coefficients.

Together with \begin{align*} \frac{(a+b)!}{a!b!}=\lim_{q\rightarrow 1}\begin{bmatrix}a+b\\a\end{bmatrix}_q \end{align*}

the claim $a!b!|(a+b)!$ follows.

Answer 5

There's a truly beautiful proof by Tim Gowers.

And a similar question has been asked before.