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Lets take the example, if we take the expression $\frac{X!}{y_1!\cdot y_2!\cdots y_n!} $as long as summation $S=y_1+y_2+...y_n$ is less than or equals $X$, the remainder is always $0$. Thats How the permutation of $X$ things where there is $y_1$ things same , $y_2$ things same works. My question is, why does this happen, what is the mathematical explanation behind this? when its like $\frac{100!}{49!\cdot49!}$ that still works? Here the first $49$ consecutive digits already divided, but how the second consecutive $1..49$ also divides by $50...100$?

paquechee
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Tamim Addari
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  • Because $n!$ divides any product of $n$ consecutive integers. – Beni Bogosel Nov 12 '13 at 19:27
  • But why, when its like 100!/(49!*49!) that still works? here the first 49 consecutive digits already divided, but How the second consecutive 1..49 also divides by 50...100? – Tamim Addari Nov 12 '13 at 19:36
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    Of course it still works. $100$ is greater than $49+49$ so you can find at least two sets of consecutive integers of length $49$ in $1,2,3,..,100$. – Beni Bogosel Nov 12 '13 at 19:38

3 Answers3

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Define $\nu_p(n)=k$, where $k$ is the power of $p$ in the prime factorisation of $n$. Evidently $\nu_p(n!)=\sum_{k \ge 1} \left \lfloor \frac{n}{p^k} \right \rfloor$ (this essentially counts the number of multiples of $p$ that are at most $n$, then double counts for multiples of $p^2$, then double counts again for multiples of $p^3$...).

Now, let's take the worse case, that $X=y_1+\cdots+y_n$, as it is true for this value of $X$ it is evidently true for larger values of $X$. We want to prove that $$\nu_p(y_1!\cdots y_n!)=\nu_p(y_1!)+\cdots+\nu_p(y_n!)\le \nu_p((y_1+ \cdots + y_n)!)$$ for an arbitrary prime $p$, so that there are no primes in the denominator that cannot be cancelled. This is equivalent to

$$\sum_{k \ge 1} \sum_{j=1}^n \left \lfloor \frac{y_n}{p^k} \right \rfloor\le \sum_{k \ge 1} \left \lfloor \frac{y_1+\cdots+y_n}{p^k} \right \rfloor.$$ This inequality evident as $$\sum_{k}\left \lfloor x_k \right \rfloor \le \left \lfloor\sum_{k} x_k \right \rfloor .$$ Here's a simple 'proof' for the above : consider $N$ blocks of wood. If you cut each block down to the nearest $\text{cm}$, the stack will be shorter than if you stacked them all up then cut the stack to the nearest $\text{cm}$.

Meow
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Not a full answer, but an outline of how I convinced myself of this fact:

Prime factorization is the key word. The key result is that if $y_1 + y_2 + \cdots y_n \leq x$ then for any prime $p$ the power of $p$ in the factorization of $x!$ is at least as high as in $y_1!y_2!\cdots y_n!$.

Once you've convinced yourself of this, you can (pretend to) write down a prime factorization of the numerator and denominator and see that the denominator can be simplified away completely. Thus you end up with just a product of primes (whatever's left of the numerator after the simplification), which is trivially an integer.

Arthur
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This is not exactly a proof. What I know from combinations is that:

$\binom{N}{m}=\frac{N!}{m!(N-m)!}$ is always integer.

That means that if N=a+b, then a!b! divides N!

You can use it more and more. Eg if also a=c+d then c!d! divides a!.

So b!c!d! divides N! where N=b+c+d. And you continue splitting.

As for why combinations is integer, there is a proof here:

Community
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Thanos Darkadakis
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