2017 AMC 12A Problem 14

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.