Question

A classroom of children has 18 boys and 19 girls in which five students are chosen to do presentations. in how many ways can the five students be chosen so that more boys than girls are selected?

Answer 1

They can be chosen in 12,444 ways.

There are three different scenarios that have more boys being chosen than girls:
5 boys and 0 girls
4 boys and 1 girl
3 boys and 2 girls

We will set up combinations for all 3 of these scenarios:
$(_{18}C_5\times{_{19}C_0)+(_{18}C_4\times_{19}C_1)+(_{18}C_3\times_{19}C_2)$
$\frac{18!}{5!13!}\times \frac{19!}{0!19!}+\frac{18!}{4!14!}\times \frac{19!}{1!18!}+\frac{18!}{3!15!}\times \frac{19!}{2!17!} \\ \\=8568+3060+816=12444$