2024 AMC 8 Problem 25

Question

A small airplane has

4

rows of seats with

3

seats in each row. Eight passengers have boarded the plane and are distributed randomly among the seats. A married couple is next to board. What is the probability there will be

2

adjacent seats in the same row for the couple?

\dfrac{8}{15}

\dfrac{32}{55}

\dfrac{20}{33}

\dfrac{34}{55}

\dfrac{8}{11}

Answer: C

Concepts:basic probability complementary counting generating functions

Difficulty rating: 1950

Small hint:

Count the four empty seats after the first eight passengers board

Big hint:

Use the complement: no row has two adjacent empty seats

Video solution:

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Written solution:

After the first

8

passengers sit, there are

4

empty seats among the

12

seats, so there are

\binom{12}{4}=495

equally likely sets of empty seats.

Count the complement, where no row has two adjacent empty seats. In a row of three seats, the possible empty-seat patterns with no adjacent empty seats have sizes

0,

1,

and

2,

with

1,

3,

and

1

choices respectively. So we need the coefficient of

x^4

in

(1+3x+x^2)^4.

This coefficient is

\begin{gathered} 3^4+4\cdot3\cdot3^2+\binom42 \\ =81+108+6 \\ =195. \end{gathered}

Therefore the probability that at least one row has adjacent empty seats is

\frac{495-195}{495}=\frac{20}{33}.

Thus, C is the correct answer.