Prove using a combinatorial arguments?

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

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  • kb
    Lv 7
    8 years ago
    Favorite 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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