2018 AMC 10B Problems/Problem 18

Three young brother-sister pairs from different families need tot take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or her sibling. How many seating arrangements are possible for this trip?

$\textbf{(A)} \text{ 60} \qquad \textbf{(B)} \text{ 72} \qquad \textbf{(C)} \text{ 92} \qquad \textbf{(D)} \text{ 96} \qquad \textbf{(E)} \text{ 120}$

Solution

We can begin to put this into cases. Let's call the pairs $a$ , $b$ and $c$ , and assume that a member of pair $a$ is sitting in the leftmost seat of the second row. We can have the following cases then.

Case 1: Second Row: a b c Third Row: b c a

Case 2: Second Row: a c b Third Row: c b a

Case 3: Second Row: a b c Third Row: c a b

Case 4: Second Row: a c b Third Row: b a c

For each of the four cases, we can flip the siblings, as they are distinct. So, each of the cases has $2 \cdot 2 \cdot 2 = 8$ possibilities. Since there are four cases, when pair $a$ has someone in the leftmost seat of the second row, there are 32 ways to rearrange it. However, someone from either pair $a$ , $b$ , or $c$ could be sitting in the leftmost seat of the second row. So, we have to multiply it by 3 to get our answer of $32 \cdot 3 = 96$ . So, the correct answer is $\boxed{D}$ .

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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2018 AMC 10B Problems/Problem 18

Solution

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