2018 AMC 10B Problem 18

P16-20~6 minPrint

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

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 without answering

Key insight

Siblings can never share a row, so each row has one child per family. Choose and order the front row; then only 2 back rows work.

Solution

Call the families $1$ , $2$ , $3$ . First find out whether two siblings can ever sit in the same row.

Step 1: siblings cannot share a row.

In a row of three seats, the only two seats that are not next to each other are the two ends. So if two siblings share a row, they take both ends, and a child from another family, say family $2$ , sits in the middle.
The other row then holds family $2$ 's other child and both children of family $3$ .
Family $3$ 's children also cannot be next to each other, so they take the two ends. That puts family $2$ 's other child in the middle seat, directly in front of or behind their sibling. Not allowed.

So each row has exactly one child from each family.

Step 2: count.

4. Who sits in front. From each family, pick which child sits in the front row: $2 \times 2 \times 2 = 8$ ways.
5. Order the front row. Three children in three seats: $3 \times 2 \times 1 = 6$ ways.
6. Order the back row. No two children in a row are siblings now, so the only rule left is that no one sits directly behind their sibling. Say the front row is $A$ , $B$ , $C$ (from families $1, 2, 3$ ), and their siblings are $a$ , $b$ , $c$ . The back row must not put $a$ in seat $1$ , $b$ in seat $2$ or $c$ in seat $3$ . List them:
- $b, c, a$ works.
- $c, a, b$ works.
- The other four orders each put someone behind their sibling.

So there are $2$ ways.
7. Multiply. $8 \times 6 \times 2 = 96$ .

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

Why this works

Checking the rules before counting showed that a whole case was impossible, which made the count simple. After that it splits into independent choices that multiply: who goes in front, their order, and a back row where nobody lands in "their own" seat. That last pattern, arranging things so none is in its matching spot, shows up often in seating problems.

Alternative approach

Count the back rows another way. Seat $a$ first: it can go in seat $2$ or seat $3$ . If $a$ takes seat $2$ , then $c$ cannot take seat $3$ , so $c$ takes seat $1$ and $b$ takes seat $3$ . If $a$ takes seat $3$ , then $b$ cannot take seat $2$ , so $b$ takes seat $1$ and $c$ takes seat $2$ . Each choice for $a$ finishes in exactly one way, so there are $2$ back rows.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $\boxed{\textbf{(D)}\ 96}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.5)
Introduction to Counting & Probability, Chapter 3: Correcting for Overcounting (§3.2, §3.4)
Prealgebra, Chapter 14: Counting (§14.3)