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

Question

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

Answer 1

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