The committee of 5 is to be chosen from 6boys and 4girls and the committe must have exactly 2 girls

## in how many ways can a committee of 5 be chosen from 6boys and 4girls

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# 1 Answer

Combinations: n!/r!(n-r)! In this case the combination of boys must be multiplied by the combination of the girls.

Boys: 6!/((6-3)!*3!) = 6*5*4*3!/3!*3! = 6*5*4/3*2*1 = 20

Girls: 4!/((4-2)!*2!) = 4*3*2!/2!*2! = 4*3/2*1 = 6

20*6 = 120

## Comments

thanks George C. this was really useful

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