AMC10 2017 A

Answer (C): Let $X$ be the set of ways to seat the five people in which Alice sits next to Bob. Let $Y$ be the set of ways to seat the five people in which Alice sits next to Carla. Let $Z$ be the set of ways to seat the five people in which Derek sits next to Eric. The required answer is $5! - |X \cup Y \cup Z|$. The Inclusion–Exclusion Principle gives $$ |X \cup Y \cup Z| = (|X| + |Y| + |Z|) - (|X \cap Y| + |X \cap Z| + |Y \cap Z|) + |X \cap Y \cap Z|. $$ Viewing Alice and Bob as a unit in which either can sit on the other’s left side shows that there are $2 \cdot 4! = 48$ elements of $X$. Similarly there are 48 elements of $Y$ and 48 elements of $Z$. Viewing Alice, Bob, and Carla as a unit with Alice in the middle shows that $|X \cap Y| = 2 \cdot 3! = 12$. Viewing Alice and Bob as a unit and Derek and Eric as a unit shows that $|X \cap Z| = 2 \cdot 2 \cdot 3! = 24$. Similarly $|Y \cap Z| = 24$. Finally, there are $2 \cdot 2 \cdot 2! = 8$ elements of $X \cap Y \cap Z$. Therefore $$ |X \cup Y \cup Z| = (48 + 48 + 48) - (12 + 24 + 24) + 8 = 92, $$ and the answer is $120 - 92 = 28$.