# Prove using a combinatorial arguments?

k*C(n,k) = n*C(n-1,k-1)

### 1 Answer

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- kbLv 78 years agoFavorite Answer
n * C(n-1, k-1)

= n * [(n-1)! / ((k-1)! (n-k)!)]

= n! / ((k-1)! (n-k)!)

= k * [n! / (k! (n-k)!)]

= k * C(n, k).

I hope this helps!

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