Explanation:
To find the number of arrangements that have at least one letter between the two Ps, we can use the concept of complementary counting.

Step 1: Total Number of Arrangements
The total number of arrangements of the letters A, P, P, L, and E is given by the formula for permutations of a set with repeated elements. Since the letter P is repeated, we have 5 letters in total, but we divide by 2! to account for the repeated P's:

Total Number of Arrangements = (5!) / (2!) = 60 / 2 = 30

Step 2: Counting the Complement
Now, we need to count the number of arrangements that do not have any letters between the two Ps. This means that the two Ps must be adjacent to each other. We can treat the two Ps as a single unit, PP, and find the number of arrangements of the letters PP, A, L, and E.

The total number of arrangements of PP, A, L, and E is given by the formula for permutations of a set with repeated elements. Since the letter P is repeated, we have 4 letters in total, but we divide by 2! to account for the repeated P's:

Number of Arrangements with PP Adjacent = (4!) / (2!) = 12 / 2 = 6

Step 3: Subtracting the Complement
To find the number of arrangements that have at least one letter between the two Ps, we subtract the number of arrangements with PP adjacent from the total number of arrangements:

Number of Arrangements with at least one letter between the two Ps = Total Number of Arrangements - Number of Arrangements with PP Adjacent
= 30 - 6
= 24

Therefore, the correct answer is option A, 24.