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.