Rather than cite the question number, please post the problem itself, along with the answer choices.
Here is the problem:
The letters D, G, I, I, and T can be used to form 5-letter strings such as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?
(A) 12
(B) 18
(C) 24
(D) 36
(E) 48
Good arrangements = total arrangements - bad arrangements.
Total arrangements:
Number of ways to arrange 5 elements = 5!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 2! to account for the two identical I's:
5!/2! = 60.
Bad arrangements:
In a bad arrangement, the two I's are in adjacent slots.
Let [II] represent the 2 adjacent I's.
Number of ways to arrange the 4 elements [II], D, G and T = 4! = 24.
Good arrangements:
Total arrangements - bad arrangements = 60-24 = 36.
The correct answer is
D.
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