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2018 AMC 10B Problem 18

2018 AMC 10B Problem 18AMC 10 Step by Step

Name: Date: September 13, 2026

Arrangements with Restrictions · difficulty 4 of 5 · about 6 min · position P16-20

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?

A)
$60$
B)
$72$
C)
$92$
D)
$96$
E)
$120$

Answer

Key insight

Siblings sharing a row forces a front-back sibling pair, so each row has one child per family: 2^3 * 3! front rows, then 2 derangements behind.

Solution

Split on whether some pair of siblings shares a row.

Case 1: two siblings sit in the same row. In a row of three seats, the only non-adjacent pair of seats is the two window seats, so the siblings take the ends and a child from a second family takes the middle. The other row then holds that child's sibling plus the complete third family. The third family must also sit at the two ends of their row, forcing the remaining child into the middle seat, directly in front of or behind their own sibling. Contradiction, so this case contributes $0$ .

Case 2: each row has one child from each family. Choose which member of each family sits in the front row: $2^3 = 8$ ways. Seat those three children in the front row: $3! = 6$ ways. Siblings are never adjacent now (no two in the same row share a family), so the only remaining rule is that nobody sits directly behind their sibling. The back row is the three remaining children, one per family, placed so that no one sits behind their own sibling: that is a derangement of $3$ positions, and there are $2$ .

Total: $8 \cdot 6 \cdot 2 = 96$ .

The answer is $\boxed{\textbf{(D)}\ 96}$ .

Why this works

Structural constraints often eliminate an entire case before any counting begins; showing that a same-row sibling pair forces a front-back pair is the real insight. After that, the count factors cleanly as (who goes in front) times (front order) times (a derangement for the back), a pattern that recurs in seating and matching problems.

Alternative approach

Count back rows for a fixed front row $F_1 F_2 F_3$ (one child per family, seats $1, 2, 3$ ). The sibling of $F_1$ may take seat $2$ or $3$ . Say seat $2$ ; then the sibling of $F_3$ cannot take seat $3$ , so it takes seat $1$ , and the sibling of $F_2$ gets seat $3$ . Each of the $2$ starting choices completes in exactly one way, so $2$ back rows per front row.

The trap

Forgetting that the back row must be a derangement (no child directly behind a sibling) and answering 8 * 6 * 6 or a similar overcount, or dismissing the case of siblings in one row without checking why it fails.

Common mistakes

Forgetting that the back row must be a derangement (no child directly behind a sibling) and answering 8 6 6 or a similar overcount, or dismissing the case of siblings in one row without checking why it fails.
Treating the two rows as interchangeable and dividing by $2$ , which gives $48$ ; the second and third rows are distinct seats.

Techniques

Split into exhaustive cases and handle each · Set up the equation/formula and compute; no special trick needed