Difference between revisions of "2018 AMC 10B Problems/Problem 18"

Revision as of 15:40, 17 December 2019

Problem

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 (Easy Casework)

Lets call the siblings $A_1$ , $A_2$ , $B_1$ , $B_2$ , $C_1$ , and $C_2$ . We can split our problem into two cases:

There is a child of each family in each row (There is an A, B, C in each row ) or There are two children of the same family in a row.

Starting off with the first case, we see that there are $3!=6$ ways to arrange the A,B,C. Then, we have to choose which sibling sits. There are $2$ choices for each set of siblings meaning we have $2^3=8$ ways to arrange that. So, there are $48$ ways to arrange the siblings in the first row. The second row is a bit easier. We see that there are $2$ ways to place the A sibling and each placement yields only $1$ possibility. So, our first case has $48\cdot2=96$ possibilities.

In our second case, there are $3$ ways to choose which set of siblings will be in the same row, two ways to choose which set of sibling will sit in between them and $2$ ways to choose whether it is the brother or sister. So, there $3*2*2 = 18$ ways to arrange the first row. In the second row, however, we see that it is impossible to make everything work out. So, there are $0$ possibilities for this case.

Thus, there are $96+0 = \boxed{D) 96}$ possibilities for this trip.

-Conantwiz2023

Solution 3

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 ( $\circ$ denotes 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 4 (a bit of group theory)

Note that there is a free $S_2\wr S_3$ 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 $S_2\wr S_3$ , which is $2^3\cdot 6=48$ . The only answer choice satisfying this is $\boxed{\textbf{(D)} \text{ 96}}$ .

To finish the solution, with a bit of work we see that there are only two orbits of seating arrangements: the orbit of 123/231 and the orbit of 123/312.

So the answer is indeed 96.

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