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Just to verify

  • #1
[tex]
\binom{n+1}{k+1}=\binom{n}{k}+\binom{n}{k+1}
[/tex]

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 [tex]\binom{p}{k}[/tex]

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)?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
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)!).
 
  • #3
[tex]
\frac{n!}{(n-k)!k!}+\frac{n!}{(n-k)!(k+1)!}
[/tex]
[tex]
\frac{n!}{(n-k-1)!(k+1)}+\frac{n!}{(n-k)!(k+1)!}
[/tex]
[tex]
\frac{(n+1)!}{(n-k)!(k+1)!}
[/tex]
[tex]
\binom{n+1}{k+1}
[/tex]
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
The first line isn't even correct. C(n,k+1)=n!/((k+1)!*(n-k-1)!).
 

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