SOLUTION: Show that (m+n)Ck = mCk * nC0 + mC(k-1) * nC1 + mC(k-2) * nC2 + ... + mC0 * nCk For k <= (m+n)

Algebra ->  Permutations -> SOLUTION: Show that (m+n)Ck = mCk * nC0 + mC(k-1) * nC1 + mC(k-2) * nC2 + ... + mC0 * nCk For k <= (m+n)      Log On


   

Question 491882: Show that (m+n)Ck = mCk * nC0 + mC(k-1) * nC1 + mC(k-2) * nC2 + ... + mC0 * nCk
For k <= (m+n)
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
Suppose Committee A has m people and Committee B has n people, and we want to choose a total of k people out of the total m+n. We can choose all k people from Committee A and none from Committee B, for mCk*nC0 ways. Or we can choose k-1 people from A and 1 person from B, and so on, hence the given statement is true.

However, this only works for k <= m and k <= n, unless we define xCy = 0, if x < y.