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?

Here's a concise explanation:

1. Identify the block structure: The problem states there are exactly two letters between A and E. This forms a block like (A _ _ E) or (E _ _ A).

2. Fill the two positions between A and E:

   • The remaining letters available are B, C, D (3 letters).
   • We need to choose 2 of these 3 letters and arrange them in the two slots.
   • Number of ways to do this = Permutations of 3 items taken 2 at a time = 3P2 = 3 * 2 = 6 ways.
   • (e.g., if the letters are B and C, they can be BC or CB).

3. Consider the order of A and E:

   • The block can be (A _ _ E) or (E _ _ A). This gives 2 possibilities.
   • So, combining with step 2, there are 6 (for the middle letters) * 2 (for A/E order) = 12 ways to form this 4-letter sequence (e.g., ABC E, EBC A).

4. Arrange the 4-letter block with the remaining letter:

   • After forming the 4-letter block (e.g., A B C E), there is 1 letter remaining (e.g., D, if B and C were chosen for the middle).
   • We now have two units to arrange: the 4-letter block and the single remaining letter.
   • These 2 units can be arranged in 2! = 2 ways (e.g., (A B C E) D or D (A B C E)).

5. Total arrangements: Multiply the possibilities from steps 3 and 4.

   • Total = 12 * 2 = 24.

Therefore, there are 24 such arrangements possible.

The final answer is C) 24.