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?

This is a problem of permutations with constraints. Let's break down the assignment process step-by-step based on the given conditions:

1. Constraint on Task-2: Task-2 must be assigned to either R or S. This gives us two main cases:

   • Case 1: R is assigned Task-2. If R gets Task-2, then R and Task-2 are fixed. Remaining persons: P, Q, S, T Remaining tasks: Task-1, Task-3, Task-4, Task-5

     Now, consider the constraint on Task-1: Neither P nor Q can be assigned Task-1. Since R is already assigned Task-2, Task-1 must be assigned to either S or T. (2 choices)

     • Subcase 1.1: S is assigned Task-1. If S gets Task-1 (and R gets Task-2), then S, R, Task-1, Task-2 are fixed. Remaining persons: P, Q, T Remaining tasks: Task-3, Task-4, Task-5 These 3 persons can be assigned to the 3 remaining tasks in 3! = 3 x 2 x 1 = 6 ways.

     • Subcase 1.2: T is assigned Task-1. If T gets Task-1 (and R gets Task-2), then T, R, Task-1, Task-2 are fixed. Remaining persons: P, Q, S Remaining tasks: Task-3, Task-4, Task-5 These 3 persons can be assigned to the 3 remaining tasks in 3! = 3 x 2 x 1 = 6 ways.

     Total ways for Case 1 (R assigned Task-2) = 6 + 6 = 12 ways.

   • Case 2: S is assigned Task-2. If S gets Task-2, then S and Task-2 are fixed. Remaining persons: P, Q, R, T Remaining tasks: Task-1, Task-3, Task-4, Task-5

     Again, consider the constraint on Task-1: Neither P nor Q can be assigned Task-1. Since S is already assigned Task-2, Task-1 must be assigned to either R or T. (2 choices)

     • Subcase 2.1: R is assigned Task-1. If R gets Task-1 (and S gets Task-2), then R, S, Task-1, Task-2 are fixed. Remaining persons: P, Q, T Remaining tasks: Task-3, Task-4, Task-5 These 3 persons can be assigned to the 3 remaining tasks in 3! = 3 x 2 x 1 = 6 ways.

     • Subcase 2.2: T is assigned Task-1. If T gets Task-1 (and S gets Task-2), then T, S, Task-1, Task-2 are fixed. Remaining persons: P, Q, R Remaining tasks: Task-3, Task-4, Task-5 These 3 persons can be assigned to the 3 remaining tasks in 3! = 3 x 2 x 1 = 6 ways.

     Total ways for Case 2 (S assigned Task-2) = 6 + 6 = 12 ways.

2. Total Ways: Sum the ways from both main cases. Total ways = 12 (from Case 1) + 12 (from Case 2) = 24 ways.

The final answer is D.