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?

The correct option is 24.

Let us find the number of possible arrangements of the five letters A, B, C, D, and E such that there are exactly two letters between A and E. First, let us analyze the positions these letters can occupy in a 5-letter sequence, numbered 1 to 5 from left to right:
_ _ _ _ _

To have exactly two letters between A and E, the positions of A and E must be separated by two index places. Thus, the only possible pairs of positions for A and E are:
1. Position 1 and Position 4
2. Position 2 and Position 5

For each of these position pairs, A and E can be arranged in two different ways (either A comes first, or E comes first):
- If the positions are 1 and 4, we can place them as (A at 1, E at 4) or (E at 1, A at 4). This gives 2 possibilities.
- If the positions are 2 and 5, we can place them as (A at 2, E at 5) or (E at 2, A at 5). This gives 2 possibilities.
Therefore, the total number of ways to position A and E is:
$2+2=4$ ways.

After placing A and E, we have 3 empty positions remaining for the other 3 letters (B, C, and D). These 3 letters can be arranged in the remaining 3 positions in 3-factorial ways:
$3!=3\times 2\times 1=6$ ways.

By the fundamental counting principle, the total number of valid arrangements is the product of the ways to place A and E and the ways to arrange the remaining letters:
$4\times 6=24$.

Thus, there are exactly 24 such arrangements possible.