The letters A, B, C, D and E are arranged in such a way that there are exactly two letters between A and E. How many such arrangements are possible?

Correct Option (C)

The problem requires arranging five distinct letters (A, B, C, D, E) such that exactly two letters are positioned between A and E.

This can be solved in two steps:

1. Determine the possible arrangements for the pair (A, E) satisfying the condition.
2. Arrange the remaining three letters (B, C, D) in the vacant positions.

For step 1, consider the five available positions (P1, P2, P3, P4, P5):

• If A is at P1, E must be at P4 (A _ _ E _).
• If A is at P2, E must be at P5 (_ A _ _ E).
• If E is at P1, A must be at P4 (E _ _ A _).
• If E is at P2, A must be at P5 (_ E _ _ A).

Thus, there are 4 distinct positional arrangements for the pair (A, E).

For step 2, after placing A and E, 3 positions remain vacant. The remaining 3 letters (B, C, D) can be arranged in these 3 positions in 3! ways.

3! = 3 × 2 × 1 = 6 ways.

The total number of such arrangements is the product of the possibilities from both steps:

Total arrangements = (Number of A-E arrangements) × (Arrangements of B, C, D)

Total arrangements = 4 × 6 = 24.

Incorrect Options:

(A) 12: This result would be obtained if only one order of A and E (e.g., A before E) was considered, leading to 2 arrangements for A and E, multiplied by 3! (2 × 6 = 12). It fails to account for the cases where E precedes A.

(B) 18: This value does not align with a standard permutation or combination calculation for this problem. It might arise from an incorrect count of A-E placements or an error in calculating permutations of the remaining letters.

(D) 36: This result could occur if the number of possible A-E arrangements was incorrectly calculated as 6 instead of 4, and then multiplied by 3! (6 × 6 = 36). An incorrect count of the initial A-E placements leads to this overestimation.