boksana wrote:

Hello!

Guys, try to solve this problem

9 people, including 3 couples are to be seated in a row of 9 chairs. What is the probability that

a. None of the Couples are together

b. Only one couple is together

c. All the couples are together

Hi, boksana!

All combinations: 9!

Combinations for which all the couples are together:

6!*2*2*2 (because there are 6! ways to arrange 6 entities - 3 couples and 3 other persons, and 2 ways of sitting a couple MW and WM).

=> p(c) = 6!*8/9! = 1/63.

Combinations for which 2 couples are together and 1 is not:

7!*2*2 - 6!*2*2*2 (the same logic - 7 entities can be seated in 7! ways, and there are 2 ways in each couple).

=> p(.) = (7!*4-6!*8)/9! = 1/18-1/63=5/126.

Combinations for which only 1 couple is:

8!*2 - [previous 2 cases together] = 8!*2 - 7!*4.

p(b) = [8!*2-7!*4]/9! = 2/9-1/18 = 1/6.

Combinations for which there are no couples:

p(a) = 1 - p(b) - p(.) - p(c) = 1 - 1/6 - 5/126 - 1/63 = 98/126 = 7/9.