just wondering if anyone knows how to do this? Show that nCk >= (n/k)^k That is, show that n chose k is greater than or equal to (n/k)^k Any ideas?

no sorry, you best bet would be to try make a triangel with 4 angles.. XD btw what kind of math is that lol?

the proof of binomial coefficient (x+y)^n = sum(k=0,n, nCk x^(n-k) y^k) and we want to show that: nCk = (n!)/(n-k)!k! is true. So let's use induction on n. Base case: n = 1. (x+y)^1 = x+y, so 1C0 = 1C1 = 1!/1!0! = 1!/0!1!. Inductive step: suppose nCk = (n!)/(n-k)!k! for some n and all 0 â‰¤ k â‰¤ n. We'll show it's true for n+1 and all 0 â‰¤ k â‰¤ n+1. So (x+y)^(n+1) = (x+y)(x+y)^n = (x+y)sum(k=0,n, nCk x^(n-k) y^k) So multiply through and note that the coefficient on x^(n+1-k) y^k is: for k = 0: 1 = (n+1)!/(n+1)!0! for k = n+1: 1 = (n+1)!/0!(n+1)! for 1 â‰¤ k â‰¤ n: nCk + nC(k-1) So we want to show that nCk + nC(k-1) = (n+1)!/(n+1-k)!k!. So: n!/(n-k!)k! + n!/(n+1-k)!(k-1)! = [n!(n+1-k)+n!k]/(n+1-k)!k! = n!(n+1)/(n+1-k)!k! = (n+1)!/(n+1-k)!k! 2+2=?+google