AMC 8 2024 Problem 25
AMC8 2024 年第 25 题
概率★★★★★
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?
一架小型飞机有4排座位，每排3个座位。目前已有8位乘客随机入座。接下来一对夫妻登机。这对夫妻能坐到同一排且相邻两个座位的概率是多少？
- A.
- B.
- C.
- D.
- E.
正确答案 · Correct Answer
C
解析 · Solution
解法1
常规解法 · 容斥（正面数「能坐」）
12 个座位坐了 8 人，剩 4 个空位；空位的位置共 种，这是分母。设 A₁…A₄ 表示「第 1…4 排出现一对相邻空位」（同排「左中」或「中右」）。要求「至少一排有」，用容斥。
第 1 步｜单排能凑出相邻空位的方法数
某排要有相邻空位，这排的空位只能是「左中」「中右」或「左中右」三种之一；剩下的空位落到另外 9 个座位上：
四排各算一遍，先相加：。
第 2 步｜扣除重叠（两排都有的情形）
第 1 步是逐排各数一次。一种坐法若两排都出现相邻空位，就会在「第 1 排有（A₁）」和「第 2 排有（A₂）」里各被数一次——重复了一次，要减回。
这种坐法的样子：4 个空位正好两对、每排一对，其余两排坐满。
它的个数 ，三个因子分别是：
· ：选哪两排（这一步只关于「排」，与夫妇无关）；
· 第一个 2：被选中的第一排那对空位是「左中」还是「中右」；
· 第二个 2：被选中的第二排那对空位是「左中」还是「中右」。
（夫妇始终不进入计数；三排都有需要 个空位，不可能。）
第 3 步｜合并
补集 · 数「坐不下」
分母同样是 。这次数反面：没有任何一排出现相邻空位。
先看单独一排（座位 ①②③），空位怎么放才没有相邻对：
· 空 0 个 —— 1 种
· 空 1 个 —— ①／②／③ 任一，3 种
· 空 2 个 —— 只能 ①③（①② 与 ②③ 都相邻），1 种
· 空 3 个 —— 必相邻，0 种
所以每排最多只能「安全」空 0、1 或 2 个。
再把 4 个空位分到 4 排（每排 0／1／2 个、合计 4），只有三种格局：
· (2,2,0,0)：选两排各空 2 个 ——
· (2,1,1,0)：空 2 的排 4 种 × 空 0 的排 3 种 × 两个「空 1」排各 3 种（）——
· (1,1,1,1)：每排空 1 个、各 3 种 ——
坐不下 =6+108+81=195。
能坐 =495-195=300，概率 。
易错提醒：
① 一排空 2 个不一定相邻——两端 ①③ 是合法的「坐不下」，漏了它补集会少算；
② 别漏 (2,2,0,0) 这种「两排都两端空」，漏了会得 （陷阱选项 D）。
【进阶】生成函数
把「一排没有相邻空位」按空位个数写成母函数 （对应空 0／1／2 个的 1、3、1 种）。四排就是 里 的系数：
再平方取 系数：
正是补集。故 。
Solution 1
Inclusion–Exclusion (count "can sit" directly)
8 of the 12 seats are taken, leaving 4 empty; the empties can be placed in ways (the denominator). Let A₁…A₄ denote "row i has an adjacent empty pair" (left–middle or middle–right). We want "at least one row," so use inclusion–exclusion.
Step 1 | one row's count
For row i to have an adjacent pair, its empty seats must be left–middle, middle–right, or left–middle–right; the remaining empties go among the other 9 seats:
Over the four rows, .
Step 2 | subtract the overlap (layouts with two such rows)
Step 1 counts each row once. A layout with adjacent pairs in two rows is counted once under "row 1 has a pair" (A₁) and once under "row 2 has a pair" (A₂) — counted twice, so subtract one copy.
What it looks like: the 4 empties form exactly two pairs, one per row, the other two rows full.
The count has three factors:
· : which two rows (about rows, not the couple);
· first 2: the first chosen row's pair is left–middle or middle–right;
· second 2: the second chosen row's pair is left–middle or middle–right.
(The couple never enters the count; three rows would need empties — impossible.)
Step 3 | combine
Complement (count "cannot sit")
Again the denominator is . Count the opposite: no row has an adjacent empty pair.
One row first (seats 1, 2, 3) — safe empty sets with no adjacent pair:
· 0 empty — 1 way
· 1 empty — seat 1, 2, or 3, so 3 ways
· 2 empty — only seats 1 and 3 (1–2 and 2–3 are adjacent), 1 way
· 3 empty — always adjacent, 0 ways
So each row may safely hold only 0, 1, or 2 empties.
Distribute 4 empties over 4 rows (each 0/1/2, total 4) — three patterns:
· (2,2,0,0): choose the two paired rows —
· (2,1,1,0): 4 for the 2-empty row × 3 for the empty row × for the two singles —
· (1,1,1,1): each row one empty —
Cannot sit =6+108+81=195.
Can sit =495-195=300, so .
Pitfalls:
1. Two empties in a row are not always adjacent — seats 1 and 3 (both ends) is a valid "cannot sit"; missing it undercounts the complement.
2. Don't drop the (2,2,0,0) case; doing so gives (trap D).
[Advanced] Generating functions
Per row, "no adjacent empties" by number of empties is (the 1, 3, 1 ways for 0/1/2 empties). Four rows means the coefficient of in :
and squaring gives the coefficient
the complement. Hence .
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