2017 AMC 12A 第 14 题

Question

Alice 拒绝坐在 Bob 或 Carla 旁边。Derek 拒绝坐在 Eric 旁边。在这些条件下，他们五个人坐成一排

5

把椅子有多少种方式？

Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of

5

chairs under these conditions?

12

16

28

32

40

从所有

5!=120

种坐法开始，减去被禁止的相邻情况

Start from all

5!=120

seatings and subtract the forbidden adjacencies

把每个被禁止的配对看作一个块（内部有

2

种顺序），并使用容斥

Treat each forbidden pair as a block (with

2

internal orders) and apply inclusion-exclusion

令

X

，

Y

，

Z

分别表示 Alice-Bob、Alice-Carla、Derek-Eric 相邻的坐法。所求为

5!-|X\cup Y\cup Z|

。

把一个被禁止的配对看作一个块，得到

|X|=|Y|=|Z|=2\cdot4!=48

。对交集，有

|X\cap Y|=2\cdot3!=12

（Alice 在 Bob 和 Carla 中间），

|X\cap Z|=|Y\cap Z|

=2\cdot2\cdot3!=24

，且

|X\cap Y\cap Z|=2\cdot2\cdot2!=8

。

由容斥，

|X\cup Y\cup Z|=(48\cdot3)

-(12+24+24)

+8=92

，所以答案是

120-92=28

。

所以正确答案是 C。

Let

X,

Y,

Z

be the seatings where Alice-Bob, Alice-Carla, and Derek-Eric are adjacent, respectively. The answer is

5!-|X\cup Y\cup Z|.

Treating a forbidden pair as a block gives

|X|=|Y|=|Z|=2\cdot4!=48.

For intersections,

|X\cap Y|=2\cdot3!=12

(Alice between Bob and Carla),

|X\cap Z|=|Y\cap Z|

=2\cdot2\cdot3!=24,

and

|X\cap Y\cap Z|=2\cdot2\cdot2!=8.

By inclusion-exclusion,

|X\cup Y\cup Z|=(48\cdot3)

-(12+24+24)

+8=92,

so the answer is

120-92=28.

Thus, the correct answer is C.

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