There are 5 tasks and 5 persons. Task-1 cannot be assigned to either person-1 or 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?

Here's a brief, clear explanation:

Let the tasks be T1, T2, T3, T4, T5 and persons be P1, P2, P3, P4, P5. Every person is assigned one task.

1. Constraint on Task-2: Task-2 must be assigned to either Person-3 or Person-4. (2 options)

2. Constraint on Task-1: Task-1 cannot be assigned to Person-1 or Person-2. Also, Task-1 cannot be assigned to the person who has already been assigned Task-2.

Let's break this down into two main cases based on Task-2's assignment:

Case 1: Task-2 is assigned to Person-3. Now, Task-1 cannot be assigned to P1, P2 (given constraint), or P3 (because P3 has T2). So, Task-1 can only be assigned to P4 or P5. (2 options) At this point, two tasks (T1, T2) are assigned to two persons (P3, and either P4 or P5). There are 3 remaining tasks (T3, T4, T5) and 3 remaining persons. These 3 tasks can be assigned to the 3 remaining persons in 3! (3 factorial) ways. 3! = 3 * 2 * 1 = 6 ways. So, for Case 1, total ways = (options for T1) * (ways for remaining tasks) = 2 * 6 = 12 ways.

Case 2: Task-2 is assigned to Person-4. Now, Task-1 cannot be assigned to P1, P2 (given constraint), or P4 (because P4 has T2). So, Task-1 can only be assigned to P3 or P5. (2 options) Similar to Case 1, there are 3 remaining tasks (T3, T4, T5) and 3 remaining persons. These can be assigned in 3! = 6 ways. So, for Case 2, total ways = (options for T1) * (ways for remaining tasks) = 2 * 6 = 12 ways.

Total number of ways = Ways for Case 1 + Ways for Case 2 = 12 + 12 = 24 ways.

The correct answer is C) 24.