2018 AMC 10B Problem 18

Problem 18 of 25IntermediateCounting & Probability

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?

Show solution

Solution

Suppose some family put both children in one row. They’d have to take the non-adjacent seats

1

and

3,

which forces the middle family’s two children into the same column. Not allowed. So each row holds exactly one child from each family. The second row is a permutation of the three families,

3! = 6

ways. The third row needs a different family in every column, a derangement of the second row’s order, and there are

2

of those. Finally, each pair can swap its two children between their seats,

2^3 = 8

ways. The total is

6 \cdot 2 \cdot 8 = 96.

Therefore, the answer is D.

Answer choices

Solution

More practice