Anonymous asked in Science & MathematicsMathematics · 4 years ago

prove nCk = n-2Ck + 2(n-2Ck-1) + n-2Ck-2 Please explain and show all steps. Thanks!?

2 Answers

  • Nick
    Lv 6
    4 years ago
    Favorite Answer

    This could be considered a double application of Pascal's rule:

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

    = C(n-2,k-2) + C(n-2,k-1) + C(n-2,k-1) + C(n-2,k)

    = C(n-2,k-2) + 2C(n-2,k-1) + C(n-2,k)

    Also it could be considered a special case of the chu-vandermonde identity with i=2:

    C(n,k) = C(n-i,k)C(i,0) + C(n-i,k-1)C(i,1) + C(n-i,k-2)C(i,2) + .. + C(n-i,k-i+2)C(i,i-2) + C(n-i,k-i+1)C(i,i-1) + C(n-i,k-i)C(i,i)

    which counts the ways of choosing k people from n-i men and i women. The left hand side counts directly the choices of k people from n and the right counts the choices of k men and 0 women, k-1 men and 1 woman, k-2 men and 2 women etc.

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