Roland2rule wrote:How many ways can the five digits 3, 3, 4, 5, 6 be arranged into a 5-digit number so that the two occurrences of the digit 3 are separated by at least one other digit?
(A) 48
(B) 36
(C) 24
(D) 18
(E) 12
Good arrangements = (all possible arrangements) - (bad arrangements).
All possible arrangements:
Number of ways to arrange 5 digits = 5! = 120.
But the five digits here include two identical 3's.
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 3's:
120/2! = 60.
Bad arrangements:
In a BAD arrangement, the two 3's are in adjacent positions.
To count the arrangements in which the two 3's are in adjacent positions, place the two 3's in a BLOCK, yielding the following list of 4 elements:
[33], 4, 5, 6
The number of ways to arrange the 4 elements {33}, 4, 5 and 6 = 4! = 24.
Good arrangements:
(all possible arrangements) - (bad arrangements) = 60 - 24 = 36.
The correct answer is
B.
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