A group of tourists consists of 3 men and 5 women. They stand in a line randomly to buy tickets to visit Angkor Wat. a) Find the number of ways the tourists can stand in a line. b) Find the probability of the following events: A: The first tourist in line is a woman. B: All the men stand next to each other.

1. Problem Description

A group of tourists consists of 3 men and 5 women. They stand in a line randomly to buy tickets to visit Angkor Wat.

a) Find the number of ways the tourists can stand in a line.

b) Find the probability of the following events:

A: The first tourist in line is a woman.

B: All the men stand next to each other.

2. Solution Steps

a) The number of ways to arrange $n$ distinct objects in a line is $n!$. In this case, there are $3 + 5 = 8$ tourists. Therefore, the number of ways to arrange them in a line is $8!$.

$8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320$

b)

A: The first tourist is a woman. The total number of tourists is

8. There are 5 women.

The number of ways to have a woman in the first position is

5. The remaining 7 tourists can be arranged in $7!$ ways.

Thus, the number of favorable outcomes is $5 \times 7! = 5 \times 5040 = 25200$.

The total number of outcomes is $8! = 40320$.

The probability of A is $P(A) = \frac{5 \times 7!}{8!} = \frac{5 \times 7!}{8 \times 7!} = \frac{5}{8}$.

B: All men stand next to each other.

Treat the 3 men as a single unit. Then there are 5 women and 1 unit of men, which makes 6 entities to arrange. These 6 can be arranged in $6!$ ways. The 3 men within the unit can be arranged in $3!$ ways.

The total number of favorable outcomes is $6! \times 3! = (6 \times 5 \times 4 \times 3 \times 2 \times 1) \times (3 \times 2 \times 1) = 720 \times 6 = 4320$.

The total number of outcomes is $8! = 40320$.

The probability of B is $P(B) = \frac{6! \times 3!}{8!} = \frac{6! \times 3!}{8 \times 7 \times 6!} = \frac{3!}{8 \times 7} = \frac{6}{56} = \frac{3}{28}$.

3. Final Answer

a) $40320$

b) $P(A) = \frac{5}{8}$ and $P(B) = \frac{3}{28}$