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Difference between revisions of "2021 April MIMC 10 Problems/Problem 16"

(Created page with "Find the number of permutations of <math>AAABBC</math> such that at exactly two <math>A</math>s are adjacent, and the <math>B</math>s are not adjacent. <math>\textbf{(A)} ~2...")
 
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==Solution==
 
==Solution==
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To be Released on April 26th.
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We can use casework counting to solve this problem.
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The first case is <math>AA\_\_\_\_</math>. Since <math>B</math> cannot be adjacent, then there are three such cases. there are <math>2!</math> for each of the case. However, <math>A</math> cannot be adjacent, therefore, there are <math>5</math> such arrangements.
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The second case is <math>\_AA\_\_\_</math>. There are total of <math>4</math> possible cases for B to not be adjacent. There are <math>1+1+1+2=5</math> total possible such arrangements. By symmetrical counting, the first case is the same as <math>\_\_\_\_AA</math> and the second case is the same as <math>\_\_\_AA\_</math>.
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The last we want to find is the number of arrangements of <math>\_\_AA\_\_</math>. For this case, there are total of <math>4</math> possible placement of two <math>B</math>s to avoid adjacency. Each has <math>1</math> arrangement. Therefore, there are total of <math>4</math> such arrangements.
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<math>5+5+5+5+4=\fbox{\textbf{(D)} 24}</math>.

Latest revision as of 12:47, 26 April 2021

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. $5+5+5+5+4=\fbox{\textbf{(D)} 24}$.

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