Question 137393
We are dealing with combinations. We don't care about the order in which the people are chosen.
For the girls n=15, r=3
15C4=n!/((n-r)!*r!)
=15!/((15-3)!*3!
=(15*14*13*12!)/(12!*3*2)
=(15*14*13)/3*2
=5*7*13
=455 combinations of 4 girls.
.
For the boys n=10, r=4
10C4=10!/((10-4)!*4!)
=(10*9*8*7*6!)/(6!*4*3*2)
=(10*9*8*7)/(4*3*2)
=5*3*2*7
=210 combinations of boys
.
455*210
=95,550 different committees can be chosen.
.
Ed