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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.

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11 Answers 11

up vote 10 down vote accepted

The key observation is that the product of $n$ consecutive integers is divisible by $n!$. This can be proved by induction.

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A quicker proof than induction would be to note that (by definition) $\binom{k}{n}$ is an integer, hence $n!|k\cdot (k-1)\ldots (k-n+2)\cdot (k-n+1)$. –  Andrew Aug 11 '10 at 17:24
Is that because given n (or more) consecutive integers one of them will always be divisible by n? –  Guillermo Phillips Aug 13 '10 at 13:05

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 $(mn)!$.

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Among many nice answers, I like this one the best. –  Pete L. Clark Aug 11 '10 at 20:17
Hmm it's also not hard to see that $\frac{(mn)!}{m!(n!)^m}$ is the number of ways of putting $mn$ distinct objects into $m$ groups, since the number of ways to permute the groups is $m!$ and the number of ways of permuting within every group is $(n!)^m$. Any not so trivial usage of that lemma? –  user21820 Dec 1 at 10:03

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 $16 < 17$. Now align the numerator and denominator terms $\rm (mod\ 16)$ by shifting the 1st numerator term so it lies above its value $\rm (mod\ 16)$, viz. $23 \equiv 7 \pmod{16}$ so right-shift the numerator terms until $23\:$ lies above $7$


Now I claim that $2$ doesn't 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 $16$ that divides $\rm d$ must also divide $\rm c$, so it cancels out upon reduction.

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 composed of the above non-aligned terms $(34 \cdots 39)/(1 \cdots 6) = \binom{39}{6}$. 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.

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As elegant as it is, this proof strikes me as requiring more mathematical sophistication than does the standard combinatorial proof of $\binom{n}{k}\in\mathbb{N}$. This makes me wonder how many "purely arithmetic" proofs of divisibility in number theory can be reduced to simpler combinatorial arguments. –  kjo Feb 1 at 21:05

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.

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While true, it's not a good answer - usually one needs to be convinced that the quotients are integers before they can understand why the results are the solution to a counting problem... –  BlueRaja - Danny Pflughoeft Aug 11 '10 at 16:19
@BlueRaja: Not at all! Showing that some ratio is the solution to a counting problem is my favourite way of proving divisibility. You don't need to be convinced that the quotient is an integer here; once you show it's the number of [something], it has to be an integer. (You don't usually check divisibility and compare prime factors when you show that the number of ways of choosing k out of n is $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.) –  ShreevatsaR Aug 11 '10 at 17:37

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):

$$\frac{9!}{2! \cdot 3! \cdot 4!}\ =\ \frac{9!}{5!\cdot 4!}\cdot \frac{5!}{2!\cdot 3!}$$

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In terms of multinomial coefficients: choosing (2, 3, and 4) out of 9 is the same as first choosing 4 from the original 9, then choosing 2 and 3 out of the remaining 5. Etc. –  ShreevatsaR Aug 11 '10 at 17:42

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.

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$$\frac{(x+y+z)!}{x!y!z!} = \binom{x + y + z}{x + y} \binom{x + y}{x}.$$

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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 $

$$ \frac{n!}{i!j!k!\cdots m!} = \binom{n}{i} \frac{(n-i)!}{j!k!\cdots m!} = \binom{n}{i}\binom{n-i}{j} \frac{(n-i-j)!}{k!\cdots m!} = \;\cdots$$

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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.

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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.

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The number of powers of $p$ dividing $n!$ is $\sum_{k=1}^n \lfloor \frac{n}{p^k} \rfloor$, and we have $\lfloor \frac{x+y+z}{q} \rfloor \geq \lfloor \frac{x}{q} \rfloor + \lfloor \frac{y}{q} \rfloor + \lfloor \frac{z}{q} \rfloor$.

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