fundamental theorem of arithmetic and moduli

shakalandro

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Nov 29, 2008
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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.
 
Its true for all n, not just primes...

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