SOLUTION: A class contains 9 boys and 11 girls. In a committee of size 5, in how many ways, girls are to be more than boys.

SOLUTION: A class contains 9 boys and 11 girls. In a committee of size 5, in how many ways, girls are to be more than boys.

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Question 1158905: A class contains 9 boys and 11 girls. In a committee of size 5, in how many ways, girls are to be more than boys.
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
committees can be 5 girls and 0 boys and that would be 11C5=462
they can be 4 girls and 1 boy and this would be 11C4*9C1=330*9=2970
or 3 girls and 2 boys and this would be 11C3*9C2=165*36=5940
Total is 9372 ways
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