# 2021 April MIMC 10 Problems/Problem 16

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Find the number of permutations of $AAABBC$ such that at exactly two $A$s are adjacent, and the $B$s are not adjacent.

$\textbf{(A)} ~21 \qquad\textbf{(B)} ~22 \qquad\textbf{(C)} ~23 \qquad\textbf{(D)} ~24 \qquad\textbf{(E)} ~25$

## Solution

We can use casework counting to solve this problem.

The first case is $AA\_\_\_\_$. Since $B$ cannot be adjacent, then there are three such cases. there are $2!$ for each of the case. However, $A$ cannot be adjacent, therefore, there are $5$ such arrangements.

The second case is $\_AA\_\_\_$. There are total of $4$ possible cases for B to not be adjacent. There are $1+1+1+2=5$ total possible such arrangements. By symmetrical counting, the first case is the same as $\_\_\_\_AA$ and the second case is the same as $\_\_\_AA\_$.

The last we want to find is the number of arrangements of $\_\_AA\_\_$. For this case, there are total of $4$ possible placement of two $B$s to avoid adjacency. Each has $1$ arrangement. Therefore, there are total of $4$ such arrangements. $5+5+5+5+4=\fbox{\textbf{(D)} 24}$.