There are 8 orators A,B,C,D,E,F,G,H. in how many ways can the arrangements be made so that A always comes before B and B before C?

• three orators A,Band C out of 8 are to be arranged in 3 out of 8 places which can be done in 8c3 ways.
  
  The remaining 5 places can be filled with remaining 5 orators in 5! ways...
  so total arrangements = = 8C3x5!
  =(8!/5!x3!)x5!
  =8!/3!
• 11 years agoHelpfull: Yes(4) No(1)
• 8 orators can be arranged in 8! ways
  
  now A, B & C can be arranged in 3! ways but we want only 1 out of these 3! ways
  
  hence the total no of ways is 8!/3!
• 11 years agoHelpfull: Yes(3) No(1)
• consider ABC as one operator as A should come before B and B From C.
  so methods of arranging 6 operators i.e. ABC,D,E,F,G,H=6!
• 11 years agoHelpfull: Yes(0) No(4)