# Thread: Prove that a variant of binomical coefficient is integer

1. ## Prove that a variant of binomical coefficient is integer

I am familiar with the standard proof that the binomial coefficient is an integer based on induction and Pascal's Rule.

I'm trying to prove that a related expression is an integer:

$\displaystyle f(m, n) = \frac{(m + n)!}{m!n!}$

Using an adaptation of Pascal's Rule, we can see that $\displaystyle f(m,n) + f(m+1,n-1) = f(m+1,n)$ and $\displaystyle f(m - 1,n) + f(m,n - 1) = f(m,n)$, but these equalities don't lead to an inductive proof.

2. ## Re: Prove that a variant of binomical coefficient is integer

Originally Posted by VinceW
I'm trying to prove that a related expression is an integer:
$\displaystyle f(m, n) = \frac{(m + n)!}{m!n!}$.
Well $\displaystyle \binom{m+n}{n} = \frac{(m + n)!}{m!n!}.$

3. ## Re: Prove that a variant of binomical coefficient is integer

Forgive me if I'm missing something, but isn't it just

$\displaystyle \frac{(m+n)!}{m! n!} = \frac{(m+n)!}{m! (m+n-m)!} = \binom{m+n}{m} \in\mathbb{N}?$

EDIT: with Plato's post above, I'm definitely not missing something!

4. ## Re: Prove that a variant of binomical coefficient is integer

Oh, yes, this is mathematically equivalent to the binomial coefficient, but I was hoping to find a more direct proof rather than converting to bionomial coefficient, and falling back on that standard proof.

5. ## Re: Prove that a variant of binomical coefficient is integer

Originally Posted by VinceW
I am familiar with the standard proof that the binomial coefficient is an integer based on induction and Pascal's Rule.

I'm trying to prove that a related expression is an integer:

$\displaystyle f(m, n) = \frac{(m + n)!}{m!n!}$

Using an adaptation of Pascal's Rule, we can see that $\displaystyle f(m,n) + f(m+1,n-1) = f(m+1,n)$ and $\displaystyle f(m - 1,n) + f(m,n - 1) = f(m,n)$, but these equalities don't lead to an inductive proof.
Hint: Try to construct a counting problem which f(m,n) is her answer.