Question

How would i solve a problem like this?

Answer 1

A committee of 5 has more boys than girls if it consists of

• 3 boys and 2 girls,

• 4 boys and 1 girl, or

• 5 boys

In the first case, there are

$\dbinom73 \dbinom62 = 525$

possible committees. In the second case,

$\dbinom74 \dbinom61 = 210$

In the third case,

$\dbinom75 \dbinom60 = 21$

So there is a total of

$\dbinom73\dbinom62 + \dbinom74\dbinom61 + \dbinom75\dbinom60 = \boxed{756}$

possible committees containing more boys than girls.