2018 AMC 10B Problems/Problem 18

Three young brother-sister pairs from different families need to 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 1 (Casework)

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 $\boxed{\textbf{(D)} \text{ 96}}$ .

Written By: Archimedes15

Solution 2

Call the siblings $A_1$ , $A_2$ , $B_1$ , $B_2$ , $C_1$ , and $C_2$ .

There are 6 choices for the child in the first seat, and it doesn't matter which one takes it, so suppose Without loss of generality that $A_1$ takes it (a $\circ$ is an empty seat):

$A_1 \circ \circ \\ \circ \ \circ \ \circ$

Then there are 4 choices for the second seat ( $B_1$ , $B_2$ , $C_1$ , or $C_2$ ). Like before, it doesn't matter who takes the seat, so WLOG suppose it is $B_1$ :

$A_1 B_1 \circ \\ \circ \ \circ \ \circ$

The last seat in the first row cannot be $A_2$ because it would be impossible to create a second row that satisfies the conditions. Therefore, it must be $C_1$ or $C_2$ . Let's say WLOG that it is $C_1$ . There are two ways to create a second row:,,,,'

$A_1 B_1 C_1 \\ B_2 C_2 A_2$

$A_1 B_1 C_1 \\ C_2 A_2 B_2$

Therefore, there are $6 \cdot 4 \cdot 2 \cdot 2= \boxed{\textbf{(D)} \text{ 96}}$ possible seating arrangements.

Written by: HoloGuard1728

Solution 3 (using the answer choices)

Note that there is a free $S_3\wr S_2$ action on the set of allowed seating arrangements: any brother-sister pair can be switched, and the 3 pairs can be permuted among each other in any way. Hence the answer must be a multiple of the order of $S_3\wr S_2$ , which is $6\cdot 2^3=48$ . The only answer choice satisfying this is $\boxed{\textbf{(D)} \text{ 96}}$ .

2018 AMC 10B (Problems • Answer Key • Resources)
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2018 AMC 10B Problems/Problem 18

Contents

Solution 1 (Casework)

Solution 2

Solution 3 (using the answer choices)

See Also