Question

Two boys and three girls will be standing in line. 1. How many ways can they line up? 2. How many ways can they line up if the boys are on the ends? 3. How many ways can they line up if the boys need to be together?

1. What is the probability the boys won't be standing together?

Answer 1

See below:

Explanation:

1

The five children can line up in any way they want. This means that we can pick one of the five children to be in spot A. And then one of the remaining four children to be in spot B. And so on. This gives:

`5xx4xx3xx2xx1=5! = 720` ways

2

If we put the boys at opposite ends, we now look at how we can arrange the boys and arrange the girls.

• Boys - we can put them on the ends in 2 different ways (boy 1 on the left or boy 2 on the left).

• Girls - we can arrange them in any which way in the middle, which is `3xx2xx1 = 3! = 6`

• Total - we now multiply the two: `2xx6=12` ways

3

If the boys have to be next to each other, now we have to figure out the number of ways:

• they can stand together (`=2`),
• the number of places in line where they can stand together (places AB, BC, CD, DE and so `=4`),
• and the number of ways the girls can be placed around them (there are 3 open spots each time, and so `3! = 6`)

This gives:

`2xx4xx6=48` ways

4

If there are 48 ways the boys can stand next to each other, that means there is `720-48=672` ways that they won't stand next to each other.

The total number of ways to arrange the five children is 720, and so the probability that the boys won't be standing next to each other is:

`672/720=14/15=0.9bar3~=93%`