2024 AMC 10A Problems/Problem 16

Problem

In how many ways can the integers $1$ , $2$ , $3$ , $4$ , $5$ , and $6$ be arranged in a line so that the following statement is true? If $2$ is not adjacent to $3$ , then $3$ is not adjacent to $4$ .

$\textbf{(A)}~480 \qquad\textbf{(B)}~504 \qquad\textbf{(C)}~528 \qquad\textbf{(D)}~572 \qquad\textbf{(E)}~600$

Solution

We take the contrapositive, where the statement becomes "If $3$ is adjacent to $4$ , then $2$ is adjacent to $3$ ."

If $3$ is adjacent to $4$ , then $2$ is also adjacent to $3$ so we can group the three numbers together in two ways: $234$ or $432$ . Then, the arrangements consists of this block of numbers and then three other numbers, which can be ordered in $4!$ ways. Hence, in this case, the total count is $2 \cdot 4! = 48$ . If $3$ is not adjacent to $4$ , there are no restrictions, so we can place $3$ and $4$ in $2 \cdot \tbinom{5}{2} = 20$ ways and then place the rest in $4! = 24$ ways, for a total of $480$ configurations.

Overall, our total count is $480 + 48 = \boxed{\textbf{(C)}~528}$ .

~joshualiu315

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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2024 AMC 10A Problems/Problem 16

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