Asked by Steve
prove (n choose k) = (n choose n-k)
Answers
Answered by
scott
it's a definition
nCk = n! / [k! * (n-k)!]
nC(n-k) = n! / {(n-k)! * [n - (n-k)]!}
n - (n-k) = k
nCk = n! / [k! * (n-k)!]
nC(n-k) = n! / {(n-k)! * [n - (n-k)]!}
n - (n-k) = k
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