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 (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 $\boxed{D}$ .

Written By: Archimedes15

Solution

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 $C_2$ ). Again, 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 $C_2$ . Suppose 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 $6 \cdot 4 \cdot 2 \cdot 2= \boxed{96}$ possible seating arrangements.

Written by: R1ceming

Solution (Using the Answers)

Notice how given an arrangement of the children that works (the answers tell us there is at least one) we can swap each pair of the siblings in one of 2 ways for $2^3$ = 8 arrangements, and each of the 3 pairs can take each others' spaces in 3! = 6 ways. This means that the answer must be divisible by 48.

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

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Solution (Casework)

Solution

Solution (Using the Answers)

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