There are 7 orators A, B, C, D, E, F and G. In how many ways can the arrangements be made so that A always comes before B and B always comes before C.	Choose the most appropriate option:
A) 7! / 3! B) 8! / 6! C) 5! x 3! D) 8! / (5! x 3!) E) 8! / (5! x 4!)
Select any three places for A, B and C. They need no arrangement amongst themselves as A would always come before B and B would always come before C.   The remaining 4 people have to be arranged in 4 places.   Thus, 7C3 x 4! = 7!x4!/(4!x3!)=7!/3!   7! / 3!NSWER:7! / 3!
<result>A</result>
There are 7 orators A, B, C, D, E, F and G. In how many ways can the arrangements be made so that A always comes before B and B always comes before C.