Question 1190992: Ten people (5 men and 5 women) are attending a dinner party. In how many ways can they be arranged if:
a. they are to be arranged in a line for a picture?
b. They are to be arranged with men and women alternating around the table?
c. they are arranged in a line for a picture but the man and women must alternate?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! **a. Arrangements in a line:**
There are 10 people in total, so there are 10! (10 factorial) ways to arrange them in a line.
10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 3,628,800
**b. Arrangements with men and women alternating around a table:**
Since there are 5 men and 5 women, they can alternate in two ways: M W M W M W M W M W or W M W M W M W M W M.
First, arrange the men. There are 5 men, so there are 5! ways to arrange them. Similarly, there are 5! ways to arrange the women.
So, the total number of arrangements is 2 × 5! × 5! = 2 × 120 × 120 = 28,800.
**c. Arrangements in a line with men and women alternating:**
Similar to the table arrangement, there are two possible alternating patterns: M W M W M W M W M W or W M W M W M W M W M.
The number of ways to arrange the men is 5! and the number of ways to arrange the women is 5!.
So, the total number of arrangements is 2 × 5! × 5! = 2 × 120 × 120 = 28,800.
**Final Answers:**
a. 3,628,800 ways
b. 28,800 ways
c. 28,800 ways
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