Q58. There are five persons, P, Q, R, S and T each one of whom has to be assigned one task. Neither P nor Q can be assigned Task-1. Task-2 must be assigned to either R or S. In how many ways can the assignment be done?

We have 5 persons (P, Q, R, S, T) and 5 tasks (Task-1 to Task-5). Each person is assigned one task, and each task is assigned to one person. Conditions: 1. Neither P nor Q can be assigned Task-1.

2. Task-2 must be assigned to either R or S. Let's break this into two cases based on Condition 2: Case 1: Task-2 is assigned to R. - Task-2: R (1 way). - Task-1: Cannot be P, Q, or R (since R is taken).

So, Task-1 can be assigned to S or T (2 ways). - Remaining 3 persons (P, Q, and the one not assigned to Task-1) can be assigned to the remaining 3 tasks in 3! = 6 ways. Total ways for Case 1 = 1 × 2 × 6 = 12 ways.

Case 2: Task-2 is assigned to S. - Task-2: S (1 way). - Task-1: Cannot be P, Q, or S (since S is taken). So, Task-1 can be assigned to R or T (2 ways). - Remaining 3 persons (P, Q, and the one not assigned to Task-1) can be assigned to the remaining 3 tasks in 3! = 6 ways.

Total ways for Case 2 = 1 × 2 × 6 = 12 ways. Total number of ways = Ways in Case 1 + Ways in Case 2 = 12 + 12 = 24 ways. This question tests permutations and combinations with multiple constraints, a common type in CSAT.