How many ways can 7 friends line up if Ann, Beth, and Chris have to stand next to each other where Ann is ahead of Beth and Beth is ahead of Chris?
Would it simply be $5*4*3*2*1=120$ ways?
Expanding on the question, say now Ann, Beth, and Chris can be in any order (but still stand next to each other). I'm not sure how to approach this, Ann can be in any of the 7 positions, but then Beth and Chris must be next to her
Treat Ann, Beth, and Christ as one block. Then we have to arrange $5$ blocks so the answer is indeed $120 = 5!$. If the order of Ann, Beth and Chris is not fixed, then the answer is $120 \cdot 3!$ from the number of ways to arrange them.