Counting in 2 ways proof

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Apr 21st, 2015

suppose we have total a+b of ball, a is white, and b is black, if we take k balls out of a+b balls and don't care about the color of the balls, the number of ways is (a+b, k).  On the other hands the k balls also can be selected by picking up i of a balls and k-i of  b balls,  which make total of k balls, i can be 0 to k, (if a>=k and b>=k), the sum of i from 0 to k, should give the total ways of selecting the balls, thus we have 

(a+b,k) = sum of i from 0 to k (a,i) (b,k-i) 

if we just define that (a,j)=0 for j>a, and (b,j)=0, for j>b    

Apr 21st, 2015

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Apr 21st, 2015
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