Question Prompt in Description below, what is the correct answer to each of the questions?

Question

Seven new employees, two of whom are married to each other, are to be assigned seven desks that are lined up in a row.

1. If the assignment of employees to desks is made randomly, what is the probability that the married couple will have adjacent desks? (Round your answer to the nearest tenth of a percent.)
2. What is the probability that the married couple will have nonadjacent desks? (Round your answer to the nearest tenth of a percent.)

Answer 1

Seven new employees, two of whom are married to each other, are to be assigned seven desks that are lined up in a row.

There are 7! = possible assignments of employees to desks (7 choices for the first desk, 6 for the next, ...)

There are 6 possible assignments of two employees to adjacent desks (1+2, 2+3, ..., 6+7)

Assign each employee a unique number from 1 to 7. Let the married employees have numbers 1 and 2.

Consider all the permutations of 1234567 and count when 12 and 21 occur together in a permutation. This has the same result as treating 12 (or 21) as a single assignment, so there are (2)6! assignments of employees to desks when counting the assignment of the married employees to two adjacent desks (in two ways) as a single assignment.

1. If the assignment of employees to desks is made randomly, what is the probability that the married couple will have adjacent desks? (Round your answer to the nearest tenth of a percent.) = (2)6!/7! = 2/7 = 28.6%
2. What is the probability that the married couple will have nonadjacent desks? (Round your answer to the nearest tenth of a percent.)

This corresponds to the permutations of 1234567 where 1 and 2 are not adjacent. There are 7 possible assignments of a desk for employee 1. Let's look at each case separately.

Employee 1 is assigned the desk in location 1: there are 5 choices for assigning employee 2 to a non-adjacent desk. There are 5! choices for assigning the remaining 5 employees to desks.

Employee 1 is assigned the desk in location 7: This is the same as case 1.

Employee 1 is assigned the desk in location 2: there are 4 choices for assigning employee 2 to a non-adjacent desk. There are 5! choices for assigning the remaining 5 employees to desks.

Employee 1 is assigned the desk in location 3,4,5 or 6: This is the same as case 2.

The total number of assignments for employees 1 and 2 not adjacent is the sum of these numbers:

2(5)(5!) + 5(4)(5!) = (30)5! out of 7! total number of random assignments. Answer = (5)6!/7! = 5/7= 71.4%.