Hard Math Problem

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 ≤ k ≤ n. We'll show it's true for n+1 and all 0 ≤ k ≤ 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 ≤ k ≤ 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

Sorry, I only know "old math" where the problems are all numbers.