All distinct arrangements of the eight letters of the word CALCULUS are written down.

The word has C, L, U twice each and A, S once, so the total is $\frac{8!}{2!2!2!}=5040$. Let $C,L,U$ denote the events that the respective pair is adjacent. Then $|C|=\frac{7!}{2!2!}=1260$ (and likewise $|L|=|U|=1260$), $|C\cap L|=\frac{6!}{2!}=360$ (same for each pair) and $|C\cap L\cap U|=5!=120$. Inclusion-exclusion gives $5040-3\cdot1260+3\cdot360-120=2220$, so (A) is true. (B) is false since $|C|=1260$. For (C), treat AUU (3 internal orders) as a block among 6 units with C, L repeated: $3\cdot\frac{6!}{2!2!}=540\ne360$. For (D), the first and last letters are a pair C, L or U ($3$ choices) and the middle six letters give $\frac{6!}{2!2!}=180$ arrangements, so $540\ne600$.