Proving Prime Divisibility of \binom{p}{k}

[chaotixmonjuish]

<br /> \binom{n+1}{k+1}=\binom{n}{k}+\binom{n}{k+1}<br />

I'm not sure how to prove this.

However...does this work:


If p is a positive prime number and 0<k<p, prove p divides \binom{p}{k}

Can't I just say that if that binomial is prime, this means that it is only divisible by p and 1 (since we are only working in the positive)?

 

[Dick]

Are these like two totally separate problems? I don't know what the second has to do with the first. The first one is just a 'find the common denominator' problem and show both sides are equal. Use C(n,k)=n!/(k!*(n-k)!).

 

[chaotixmonjuish]

<br /> \frac{n!}{(n-k)!k!}+\frac{n!}{(n-k)!(k+1)!}<br />
<br /> \frac{n!}{(n-k-1)!(k+1)}+\frac{n!}{(n-k)!(k+1)!}<br />
<br /> \frac{(n+1)!}{(n-k)!(k+1)!}<br />
<br /> \binom{n+1}{k+1}<br />

 

[Dick]

The first line isn't even correct. C(n,k+1)=n!/((k+1)!*(n-k-1)!).