2017 AMC 12A Problem 14

Question

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

Start from all

5!=120

seatings and subtract the forbidden adjacencies

Treat each forbidden pair as a block (with

2

internal orders) and apply inclusion-exclusion

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.

Problem 14 in other years