We are given that there are 4 boys and 5 girls standing in a line. We are asked to find: a) The total number of ways they can stand in a line. b) The number of ways they can stand in a line if: a) The first 2 people are girls. b) The first person is a boy and the last person is a girl. c) The boys stand together. d) No two girls stand next to each other.

1. Problem Description

We are given that there are 4 boys and 5 girls standing in a line. We are asked to find:

a) The total number of ways they can stand in a line.

b) The number of ways they can stand in a line if:

a) The first 2 people are girls.

b) The first person is a boy and the last person is a girl.

c) The boys stand together.

d) No two girls stand next to each other.

2. Solution Steps

a) Total number of ways they can stand in a line.

There are a total of $4+5=9$ people. The number of ways to arrange 9 people is $9!$.

$9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362880$

b)

a) The first 2 people are girls.

There are 5 girls. For the first position, there are 5 choices. For the second position, there are 4 choices.

The remaining 7 people can be arranged in $7!$ ways.

So the total number of ways is $5 \times 4 \times 7! = 20 \times 5040 = 100800$.

$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$

b) The first person is a boy and the last person is a girl.

There are 4 choices for the first position (a boy) and 5 choices for the last position (a girl). The remaining 7 people can be arranged in $7!$ ways.

So the total number of ways is $4 \times 5 \times 7! = 20 \times 5040 = 100800$.

c) The boys stand together.

Consider the 4 boys as a single unit. Then we have 5 girls and 1 unit of boys, which is a total of 6 units. These 6 units can be arranged in $6!$ ways. The 4 boys can be arranged within their unit in $4!$ ways.

So the total number of ways is $6! \times 4! = (720) \times (24) = 17280$.

$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$

$4! = 4 \times 3 \times 2 \times 1 = 24$

d) No two girls stand next to each other.

Since there are 4 boys and 5 girls, it is necessary that at least two girls stand next to each other. Thus, there are 0 ways that no two girls stand next to each other.

3. Final Answer

a) $9! = 362880$

b)

a) $5 \times 4 \times 7! = 100800$

b) $4 \times 5 \times 7! = 100800$

c) $6! \times 4! = 17280$

d) $0$