There are 10 orators A, B, C, D, E, F, G, H, I and J. In how many ways can the arrangements be made so that A always comes before B and B always comes before C.	Options:  A) 10! / 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 7 people have to be arranged in 7 places.   Thus, 10C3 x 7! =10! / 3!   10! / 3!NSWER:10! / 3!
<result>A</result>
There are 10 orators A, B, C, D, E, F, G, H, I and J. In how many ways can the arrangements be made so that A always comes before B and B always comes before C.