Division of Factorials [binomal coefficients are integers] I have a partition of a positive integer $(p)$. How can I prove that the factorial of $p$ can always be divided by the product of the factorials of the parts?
As a quick example $\frac{9!}{(2!3!4!)} = 1260$ (no remainder), where $9=2+3+4$.
I can nearly see it by looking at factors, but I can't see a way to guarantee it.
 A: Below is a sketch of a little-known purely arithmetical proof that binomial coefficients are integral. I purposely constructed the proof so that it would be comprehensible to an educated layperson. The proof gives an algorithm to rewrite a binomial coefficient as a product of fractions whose denominators are coprime to any given prime. The method of proof is  best comprehended by applying the algorithm to a specific example [Note: It may prove helpful to first read this simpler example before proceeding to the exposition below].
E.g. $ $ consider $\  \binom{39}{17}\, =\, \frac{39!}{22!\, 17!}\, =\, \frac{23 \cdot 24 \cdots\; 39}{1 \cdot 2 \cdots\; 17}.\, $ When this fraction is reduced to lowest terms $\rm\:a/b,\, $ no prime $\rm\ p > 17\ $ can divide its denominator $\rm\: b,\: $ since $\rm\ b\:|\:17\:!\ \:$ Hence, to show that $\rm\ a/b\ $ is an integer, it suffices to show that no prime $\rm\ p \le 17\ $ divides its denominator $\rm\: b$.
E.g. we show that $\,2\,$ doesn't divide $\rm b$. The highest power of $\,2\,$ in the
denominator terms is $16 < 17$. Align the numerators & denominators $\!\bmod 16\,$ by shifting the 1st numerator term so it lies above its value $\!\bmod 16,\,$
viz. $\,\color{#c00}{23 \equiv 7} \pmod{\!16}\,$ so right-shift the numerator terms until $\,23\,$ lies above $\:\!7\,$
$$\color{#0a0}{\frac{}{1}\frac{}{2}\frac{}{3}\frac{}{4}\frac{}{5}\frac{}{6}}\color{#c00}{\frac{23}{7}}\frac{24}{8}\frac{25}{9}\frac{26}{10}\frac{27}{11}\frac{28}{12}\frac{29}{13}\frac{30}{14}\frac{31}{15}\frac{32}{16}\frac{33}{17}\color{#0a0}{\frac{34}{}\frac{35}{}\frac{36}{}\frac{37}{}\frac{38}{}\frac{39}{}}$$
We claim that $\,2\,$ does not divide the reduced denominator of each aligned fraction. Indeed $\ 24/8 = 3$, $\: 26/10 = 13/5 $,  $\:\ 28/12 = 7/3 $,  $\:\ 30/14 = 15/7 $,  $\:\ 32/16 = 2$.
This holds because these fractions $\rm\; c/d \;$  satisfy  $\,\rm c \equiv d\ (mod\ 16)\:$
i.e.  $\,\rm c = d + 16\:\! n \;$  so  $\,\rm 2|d \Rightarrow 2|c$, $\rm\; 4|d \Rightarrow 4|c$, $\:\cdots $, $\rm\; 16|d \Rightarrow 16|c,\,$
i.e. any power of $2\:$ below $16$ dividing $\rm d$ must divide $\rm c$, so it cancels out of $\rm\,c/d$.
Therefore to prove that $2$ doesn't divide the reduced denominator of $\binom{39}{17}$
it suffices to prove the same for the "leftover" fraction $\,\color{#0a0}{(34 \cdots 39)/(1 \cdots 6) = \binom{39}{6}}\,$ composed
of the above non-aligned terms. Being an $\rm\binom{n}{k}$ with smaller $\rm k = 6 < 17,\,$ this follows by induction.
Since the same proof works for any prime $\rm p$, we conclude that no prime divides the
reduced denominator of $\,\binom{39}{17},\,$ therefore it is an integer.$\quad$ QED
Informally, the reason that this works is because the denominator sequence starts at $1$, which is coprime to every prime $\rm p$. This ensures that it is the "greediest" possible contiguous sequence of integers, in the sense that its product contains the least power of $\rm\:p\:$ compared to other contiguous sequences of equal length.
The algorithm extends to multinomials by using the simple reduction of multinomials to products of binomials mentioned in my prior post here.
A: To show that $\frac{(x+y+z)!}{x!y!z!}$ is an integer (when $x$, $y$, and $z$ are non-negative integers) it is enough to show that $\frac{(x+y+z)!}{x!y!z!}$ counts something. 
Consider all words of length $x+y+z$ made up of $x$ occurrences of the letter A, $y$ occurrences of the letter B, and $z$ occurrences of the letter C. There are $\frac{(x+y+z)!}{x!y!z!}$ such words. 
A: The "high-level" way to see this is to recall that whenever a finite group $G$ has a subgroup $H$, we know that $|H|$ divides $|G|$.  Then note that $S_n$ clearly contains $S_{n_1} \times ... \times S_{n_k}$ as a subgroup for any partition $n_1 + ... + n_k = n$.  (This is actually the same as the combinatorial interpretation in Robin Chapman's answer, since what we are counting is the number of cosets $G/H$, and these cosets are precisely what the multinomial theorem is counting.)
This basic lemma is surprisingly useful.  For example, it is not hard to use it to show that $m! (n!)^m$ divides $(mn)!$.
A: Let's prove using induction, the special case of two numbers, i.e., the statement that if $p, q$ $\in \mathbb{N}$ then $p!q!|(p+q)!$.
(Assume any new variables introduced below to refer to natural numbers.)
First note that since $(p+1)! = (p+1)p!1!$, the statement is true for $q = 1$ and any $p$ (including $p = 1$). In particular, it is true for $p + q = 1 + 1 = 2$. Let us assume that the statement holds for $p, q$ such that $p + q = n$.
Now, for $p, q$ such that $p + q = n + 1$, we can write 
$$(p + q)! = (p + q)(p + q - 1)!$$ 
$$= p [\underbrace{(p - 1) + q}_n]! + q [\underbrace{p + (q - 1)}_n]!$$ 
$$= p \underbrace{k_1 (p-1)!q!}_{\text{using induction assumption}} + q \underbrace{k_2 p!(q-1)!}_{\text{using induction assumption}}$$ 
$$= (k_1 + k_2) p! q!$$. Hence the principle of mathematical induction implies the truth of the statement.
Now it is easy to prove the analogous statement for three numbers, i.e. $p!q!r! | (p + q + r)!$, since (using the statement just proven) $(p + q + r)!$ is divisible by $p! (q+r)!$ and $(q + r)!$ is divisible by $q!r!$.
This can be generalised for any number of parts, by induction.
A: Here is another proof:
We use Legendre's formula for the exact power of a prime $p$ which divides $n!$ which is given by
$$ \sum_{k=1}^{\infty} \left\lfloor\frac{n}{p^k}\right\rfloor$$
where $\lfloor x\rfloor$ is the integer part of $x$.
Coupled with $$ \sum_{i=1}^{n}  \left\lfloor x_i\right\rfloor \le \left\lfloor\sum_{i=1}^{n} x_i \right\rfloor$$ 
we get that if $\sum_{i=1}^{n} a_i = N$ then
$$\sum_{i=1}^{n} \left\lfloor\frac{a_i}{p^k}\right\rfloor \le \left\lfloor\frac{N}{p^k}\right\rfloor $$
And so any prime power which divides $(a_1)! \dots (a_n)!$ divides $N!$ and so $\displaystyle \frac{N!}{(a_1)! \dots (a_n)!}$ is an integer.
A: These quotients are integers since they solve counting problems.
For instance, how many nine-letter words are there with 2 As, 3 Bs and 4 Cs?
For the full story see the multinomial theorem.
A: If you believe (:-) in the two-part Newton case, then the rest is easily obtained by induction. For instance (to motivate you to write a full proof):
$$\frac{9!}{2! \cdot 3! \cdot 4!}\ =\ \frac{9!}{5!\cdot 4!}\cdot \frac{5!}{2!\cdot 3!}$$
A: The key observation is that the product of $n$ consecutive integers is divisible by $n!$. This can be proved by induction.
A: Besides the obvious combinatorial interpretations, one also has the has the following reduction from multinomial coefficient to products of binomial coefficients. Namely for $n = i+j+k+\cdots + m $
$$ \frac{n!}{i!j!k!\cdots m!} = \binom{n}{i} \frac{(n-i)!}{j!k!\cdots m!} = \binom{n}{i}\binom{n-i}{j} \frac{(n-i-j)!}{k!\cdots m!} = \;\cdots$$
A: $$\frac{(x+y+z)!}{x!y!z!} = \binom{x + y + z}{x + y} \binom{x + y}{x}.$$
A: Hint.  Write
$$\eqalign{(x+y+z)!
  &=[(1)(2)\cdots(x)]\cr
  &\qquad{}\times[(x+1)(x+2)\cdots(x+y)]\cr
  &\qquad{}\times[(x+y+1)(x+y+2)\cdots(x+y+z)]\cr}$$
and use a well-known fact about a product of $n$ consecutive integers.
A: The number of powers of $p$ dividing $n!$ is $\sum_{k=1}^n \lfloor \frac{n}{p^k} \rfloor$, and we have $\lfloor \frac{x+y+z}{q} \rfloor \geq \lfloor \frac{x}{q} \rfloor + \lfloor \frac{y}{q} \rfloor + \lfloor \frac{z}{q} \rfloor$.
