Question

In how many ways can 2 men and 3 women sit on a straight line if the men must sit on the ends ?

Answer 1

2 ways to seat the men on the ends, 6 ways to seat the women in the middle, giving `2xx6=12` ways.

Explanation:

We have 5 seats, 2 men, 3 women, and men must sit on the ends.

Let's talk about the ends first.

There are 2 men and 2 places where they can sit. This is a permutation problem (in that person A sitting in seat 1 and person B sitting in seat 5 is different than A sitting in 5 and B sitting in 1). The general formula for a permutation is:

`P_(n,k)=(n!)/((n-k)!); n="population", k="picks"`

`P_(2,2)=(2!)/((2-2)!)=(2!)/(0!) =2/1=2`

We now have 3 seats in the middle and 3 women who can sit there (the remaining man and 2 women). We get:

`P_(3,3)=(3!)/((3-3)!)=(3!)/(0!) =6/1=6`

And now we multiply the two numbers together:

`2xx6=12`