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:
$(_{18}C_5\times{_{19}C_0)+(_{18}C_4\times_{19}C_1)+(_{18}C_3\times_{19}C_2)$
$\frac{18!}{5!13!}\times \frac{19!}{0!19!}+\frac{18!}{4!14!}\times \frac{19!}{1!18!}+\frac{18!}{3!15!}\times \frac{19!}{2!17!} \\ \\=8568+3060+816=12444$

Answer 2

Using the combination principle, the number of ways of selecting 5 students such that the number of boys is greater the number of girls selected ls 206244 ways.

Number of boys = 18

Number of girls = 19

Number of selections to be made = 5

In other to choose more boys than girls :

• 5 boys and 0 girls
• 4 boys and 1 girl
• 3 boys and 2 girls

Using the combination relation :

(18C5 × 19C0) + (18C4 × 19C1) + (18C3 × 19C2)

(8568 × 1) + (3060 × 19) + (816 × 171)

(8568 + 58140 + 139536)

= 206244 ways

Therefore, the Number of ways of making the selection is 206244 ways

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