Posted by **Stephen** on Friday, October 3, 2008 at 8:04pm.

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**, Friday, October 3, 2008 at 9:05pm
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

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