Problem 91 - Algebra and Trigonometry

Question

Five men and five women line up at a checkout counter in a store. In how many ways can they line up if the first person in line is a woman and the people in line alternate woman, man, woman, man, and so on?

Step-by-Step Solution

Answer

The total number of arrangements is 14400.

Step 1: Arrange women first

We arrange the five women first, who must begin the line. The number of ways these five women can arrange themselves in five spots is given by permutation of 5 elements, which is noted by the formula \(P(n,r) = \frac{n!}{(n-r)!}\) where \(n\) is the total number of items, and \(r\) is the items to choose from. In this case, \(n = r = 5\), therefore it yields \(P(5,5) = \frac{5!}{(5-5)!}\) which is equal to 120.

Step 2: Arrange men

Now we arrange the five men in the five remaining spots. Similar to the women, the number of ways these five men can arrange themselves in five spots can also be calculated by permutation of 5 elements, which is \(P(5,5) = \frac{5!}{(5-5)!}\) = 120.

Step 3: Total arrangements

Because the arrangements of men and the arrangements of women can be done independently, we multiply the numbers of arrangements to get the total number of ways they can line up with the constraints of the problem. Therefore, the total number of ways equals to 120 * 120 = 14400.