There are 6 tasks and 6 persons. . . . .

To add on to what regor60 said above — the reason we start by assigning task 2 is because it's the most restrictive step. It's challenging to start with task 1, for instance, because the assignments for task 2 limit the options for task 1. After assigning task 2, we move to task 1 because it's the second most restrictive step. The remaining tasks have no restrictions, so we can assign them in any order.

So we have

Task 2: 2 options (P3 or P4)
Task 1: 3 options (P3/P4 [depending on assignment of Task 2] or P5 or P6)
Task 3: 4 options (remaining four people)
Task 4: 3 options (remaining three people)
Task 5: 2 options (remaining two people)
Task 6: 1 option (remaining one person)

2*3*4*3*2*1 = 144

Vincen wrote: ↑

Fri Nov 10, 2017 6:06 am

There are 6 tasks and 6 persons. Task 1 cannot be assigned either to person 1 or to person 2; Task 2 must be assigned to either person 3 or person 4. Every person is to be assigned one task. In how many ways can the assignment be done?

(1) 144

(2) 180

(3) 192

(4) 360

(5) 716

The OA is A .

Experts, what is the formula that I should apply here?

Solution:

Case 1. Task 2 is assigned to person 3.

If task 2 is assigned to person 3, then there are 3 people who can perform task 1 (person 4, 5, or 6, since person 1 and 2 cannot perform task 1, and person 3 will be assigned to task 2). Then, for tasks 3 through 6, inclusive, there are no restrictions. Thus, there are 3 x 1 x 4 x 3 x 2 x 1 = 72 ways for the assignment to be done.

Case 2. Task 2 is assigned to person 4.

This situation is identical to Case 1, except that person 4 (vice person 3) is assigned to task 2. Thus, the number of ways for the assignment is the same as for Case 1, which is 72.

Thus, there are 72 + 72 = 144 ways to make the assignment.

Answer: A