Probability Question

Question

(Seating arrangements.) In how many ways can 8 people be seated in a Round table if (a) there are no restrictions on the seating arrangement? (b) persons A and B must sit next to each other? (c) there are 4 men and 4 women, and no 2 men or 2 women can sit next to each other? (d) there are 5 men and they must sit next to each other? (e) there are 4 married couples and each couple must sit together?

Answer 1

We are assuming that it matters which chair someone sits in (thus, everybody moving over one is not the same arrangement).

Note that there are many ways that these problems can be solved. I could solve some of these by using fractions of the total ways that could follow a particular order, but I will use only multiplications here to be consistent.

a) 8! = 40320

b) Once person A sits, there are 2 possible seats for person B. Then, the remaining 6 can sit anywhere. 8 * 2 * 6! = 11520

c) Since men and women must alternate, there can be either a man or woman in seat 1. Then, there are 4! ways to seat people from the same sex in 1, 3, 5, and 7, and 4! ways to seat people from the same sex in seats 2, 4, 6, and 8

2 * (4!)^2 = 2 * 24^2 = 1152

d) As the 5 men must sit together, there are 8 different chairs where the first man can sit, 5! orders of men, and 3! orders of women, so 8 * 5! * 3! = 8 * 120 * 6 = 5760

e) There are two possible spots where the couples can start seating. Then, there are 4! orders of the couples, and 2^4 orders within the couples. 2 * 4! * 2^4 = 2 * 24 * 16 = 768