IAS 2023 — CSAT Reasoning

2023mcqmedium

There are five person 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?

Official previous-year question

Verified 2 Aug 2026.

Options

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Solution

Correct Option (4)

The problem requires assigning five persons (P, Q, R, S, T) to five distinct tasks, with the following constraints:

P cannot be assigned Task-1.
Q cannot be assigned Task-1.
Task-2 must be assigned to either R or S.

We approach this by considering two mutually exclusive cases based on the assignment of Task-2:

Case 1: Task-2 is assigned to R.

Task-2 is assigned to R. This accounts for 1 way.
For Task-1:
- It cannot be assigned to P or Q (as per the given conditions).
- It cannot be assigned to R (as R has already been assigned Task-2).
- Therefore, Task-1 can only be assigned to either S or T. This provides 2 choices.
After Task-2 and Task-1 are assigned, 3 persons and 3 tasks (Task-3, Task-4, Task-5) remain. These 3 remaining persons can be assigned the 3 remaining tasks in 3! (3 factorial) ways.
- 3! = 3 × 2 × 1 = 6 ways.
The total number of ways for Case 1 is the product of the choices at each step: 1 (for Task-2) × 2 (for Task-1) × 6 (for remaining tasks) = 12 ways.

Case 2: Task-2 is assigned to S.

Task-2 is assigned to S. This accounts for 1 way.
For Task-1:
- It cannot be assigned to P or Q (as per the given conditions).
- It cannot be assigned to S (as S has already been assigned Task-2).
- Therefore, Task-1 can only be assigned to either R or T. This provides 2 choices.
After Task-2 and Task-1 are assigned, 3 persons and 3 tasks (Task-3, Task-4, Task-5) remain. These 3 remaining persons can be assigned the 3 remaining tasks in 3! ways.
- 3! = 3 × 2 × 1 = 6 ways.
The total number of ways for Case 2 is the product of the choices at each step: 1 (for Task-2) × 2 (for Task-1) × 6 (for remaining tasks) = 12 ways.

Since Case 1 and Case 2 are mutually exclusive, the total number of ways to assign the tasks is the sum of the ways from both cases.

Total ways = 12 (from Case 1) + 12 (from Case 2) = 24 ways.

Incorrect Options:

Options 1 (6), 2 (12), and 3 (18) are incorrect because they represent incomplete calculations or partial scenarios, not encompassing all valid permutations under the specified conditions. The comprehensive analysis demonstrates that 24 is the correct total number of assignment possibilities.

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Correct Option (4)

The problem requires assigning five persons (P, Q, R, S, T) to five distinct tasks, with the following constraints:

P cannot be assigned Task-1.
Q cannot be assigned Task-1.
Task-2 must be assigned to either R or S.

We approach this by considering two mutually exclusive cases based on the assignment of Task-2:

Case 1: Task-2 is assigned to R.

Task-2 is assigned to R. This accounts for 1 way.
For Task-1:
- It cannot be assigned to P or Q (as per the given conditions).
- It cannot be assigned to R (as R has already been assigned Task-2).
- Therefore, Task-1 can only be assigned to either S or T. This provides 2 choices.
After Task-2 and Task-1 are assigned, 3 persons and 3 tasks (Task-3, Task-4, Task-5) remain. These 3 remaining persons can be assigned the 3 remaining tasks in 3! (3 factorial) ways.
- 3! = 3 × 2 × 1 = 6 ways.
The total number of ways for Case 1 is the product of the choices at each step: 1 (for Task-2) × 2 (for Task-1) × 6 (for remaining tasks) = 12 ways.

Case 2: Task-2 is assigned to S.

Task-2 is assigned to S. This accounts for 1 way.
For Task-1:
- It cannot be assigned to P or Q (as per the given conditions).
- It cannot be assigned to S (as S has already been assigned Task-2).
- Therefore, Task-1 can only be assigned to either R or T. This provides 2 choices.
After Task-2 and Task-1 are assigned, 3 persons and 3 tasks (Task-3, Task-4, Task-5) remain. These 3 remaining persons can be assigned the 3 remaining tasks in 3! ways.
- 3! = 3 × 2 × 1 = 6 ways.
The total number of ways for Case 2 is the product of the choices at each step: 1 (for Task-2) × 2 (for Task-1) × 6 (for remaining tasks) = 12 ways.

Since Case 1 and Case 2 are mutually exclusive, the total number of ways to assign the tasks is the sum of the ways from both cases.

Total ways = 12 (from Case 1) + 12 (from Case 2) = 24 ways.

IAS 2023 — CSAT Reasoning

Options

Did you get this right?

Solution

Correct Option (4)

Case 1: Task-2 is assigned to R.

Case 2: Task-2 is assigned to S.

Incorrect Options:

Your note

IAS Reasoning in other years

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Other CSAT topics for IAS

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IAS 2023 — CSAT Reasoning

Options

Did you get this right?

Solution

Correct Option (4)

Case 1: Task-2 is assigned to R.

Case 2: Task-2 is assigned to S.

Incorrect Options:

Your note

IAS Reasoning in other years

More IAS Reasoning questions

Other CSAT topics for IAS

Comprehension

Maths

Keep practicing Reasoning