Four couples have reserved seats in a row for a concert. How many ways can they be seated if the two members of each couple wish to sit together?
How do I go about doing this problem? I'm not sure what/how I should multiply this out...
To solve this problem, you can break it down into smaller steps and then multiply the results together.
Step 1: Consider each couple as a single entity.
Since each couple wants to sit together, you can think of each couple as one unit. So you have four units (couples) to place in a row. There are 4! (4 factorial) ways to arrange these units.
Step 2: Arrange the members within each couple.
Within each couple, you have two members who can be arranged in 2! ways. Since there are four couples, you need to account for this arrangement for each couple. So you need to multiply the 2! for each couple, four times.
Step 3: Multiply the results from Step 1 and Step 2 together.
Multiply the number of ways to arrange the units (4!) by the number of ways to arrange the members within each couple (2! x 2! x 2! x 2!).
Therefore, the total number of ways the four couples can be seated is: 4! x (2! x 2! x 2! x 2!) = 4! x 2^4
Simplifying this expression, 4! (4 factorial) means 4 x 3 x 2 x 1, and 2^4 means 2 x 2 x 2 x 2. So the total number of ways the four couples can be seated is: 24 x 16 = 384.