2018 AMC 10B Problem 18

Question

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?

60

72

92

96

120

Answer: D

Concepts:casework permutations derangements

Difficulty rating: 1930

Small hint:

Show that each row must contain exactly one child from each family

Big hint:

The third row’s families must differ from the second row’s in every column (a derangement), and each pair’s two children can be swapped between their seats

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.