fundamental theorem of arithmetic and moduli

[shakalandro]

How would I got about proving that p! = i!(p-i)!(p choose i) for any i and where p is prime. I'm suppose to do it using the fundamental theorem of arithmetic, i.e. the idea that any number is the product of primes.

 

[daon]

Its true for all n, not just primes...

\(\displaystyle k!(n-k)!{n \choose k} = k!(n-k)!\frac{n!}{(n-k)!k!}=n! \frac{k!(n-k)!}{k!(n-k)!}=n!\)