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

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- NickLv 64 years agoFavorite 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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