Posted by Stephen on .
A class has 10 boys and 12 girls. In how many ways can a committee of four be selected if the committee can have at most two girls?

Math 117 
Damon,
With no girls
C(10,4) = 210 ways
With one girl
12 C(10,3) = 12*120 = 1440 ways
With two girls
first combinations of 12 girls taken 2 at a time = 12!/[10!*2!) = 12*11/2 = 66
(my table of binomial coefficients only goes to n = 10)
time combinations of ten boys taken 2 at a time = C(10,2) = 45
so 66*45 = 2970
so
2970+1440+210