Problem 5

A group of 5 men and 3 women are to be seated around a circular table. In how many different ways can they be arranged so that no women are seated next to one another?

[Problem submitted by Roger Wolf, LACC Chairman of Mathematics.]

Solution for Problem 5:

Let A, B, and C be the 3 women. We can seat A in any one of the 8 seats. Assume A is in seat #1, then B may only take #3 through #7. If B takes #3, then C may only take #5, 6, or 7. We use the notation “A1B3C5” to stand for “A takes #1, B takes #3, and C takes #5 seat.”

The following are all the possible outcomes when A takes #1 seat:

A1B3C5, A1B3C6, A1B3C7

A1B4C6, A1B4C7

A1B5C3, A1B5C7

A1B6C3, A1B6C4

A1B7C3, A1B7C4, A1B7C5.

There are 12 different ways to seat A, B, and C when A is seated in #1. Similarly there are 12 ways to seat them when A is in #2 through 7. Therefore there are 8 * 12 ways to seat the women.

After the women are seated, there are 5! ways to seat the 5 men. Therefore the total number of ways to seat 3 women and 5 men is 8*12*5! =11,520.