There are 9 orators A, B, C, D, E, F, G,H and I. In how many ways can the arrangements be made so that A always comes before B and B always comes before C.  Options.
A) 9! / 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 6 people have to be arranged in 6 places.   Thus, 9C3 x 6! = 9! / 3!   9! / 3!NSWER:9! / 3!
<result>A</result>
There are 9 orators A, B, C, D, E, F, G,H and I. In how many ways can the arrangements be made so that A always comes before B and B always comes before C.