Question

Probability

13. Four men and three women are to be seated in a seven-chair row. Find the probability for each arrangement if a. the women will sit together b. the men and women will sit alternately. c. a man will sit in the first seat. d. the men will sit together.

Answer 1

a.

Number of ways seven people can sit on chair = 7! = 5040

Let us group all three women as one. Then we have 1 group + 4 men = 5 to be arranged in 5 positions.

Number of ways women will sit together = Number of ways three women sit on 3 chairs in the group * Number of ways to arrange in 5 positions = 3! * 5! = 6 * 120 = 720

Probability that women will sit together = 720 / 5040 = 0.1429

b.

Number of ways men and women will sit alternately = Number of ways four men sit on alternate 4 chairs * Number of ways three women sit on 3 chairs = 4! * 3! = 24 * 6 = 144

Probability that men and women will sit alternately = 144 / 5040 = 0.0286

c.

If a man will sit in the first seat, then there are 6 people to be arranged in 6 chairs = 6! = 720

Probability that man will sit in the first seat = 720 / 5040 = 0.1429

d.

Let us group all four men as one. Then we have 1 group + 3 women = 4 to be arranged in 4 positions.

Number of ways men will sit together = Number of ways four men sit on 4 chairs in the group * Number of ways to arrange in 4 positions = 4! * 4! = 24 * 24 = 576

Probability that women will sit together = 576 / 5040 = 0.1143