Seven speakers A1, A2, A3,……, A7 were scheduled to speak at a function. In how many ways their speech can be arranged, such that:
A1 speaks before A3 , and
A3 speaks before A5 .

• Ans: 840
  
  In these type of problems consider ordered elements as say ($)
  A1,A2,A3,A4,A5,A6,A7
  Now consider the ordering and replace them with our assumed symbol ($)
  $,A2,$,A4,$,A6,A7
  So, its nothing but arrangement of n items out of which m items are similar. Which is given by n!/m!
  Therefore, arrangement of 7 items out of which 3 items are similar --> 7!/3! =840
• 11 years agoHelpfull: Yes(22) No(15)
• A1,A3,A5 as a group say A
  along with remaining A2,A4,A6,A7 can be arranged in 5! ways,in a group only 1 order A1 A3 A5
  so ans is 5!*1!=120
• 11 years agoHelpfull: Yes(16) No(4)
• 5!
  is answer
  
• 11 years agoHelpfull: Yes(11) No(3)
• A3 should be between A1 and A5
  so if we fix the position of A1 and A5 in the beginning and at the end respectively we get A1 A2 A3 A4 A6 A7 A5, we can arrange A2 A3 A4 A6 A7 in 5! ways
  which will satisfy the condition. There are 5! ways=120
• 11 years agoHelpfull: Yes(5) No(0)
• choose any 3 places from 7 7c3 ...do not arrange them as....we want only 1 pattern...i.e. a1,a3,a5....and rest 4 places can be arranged in 4! ways...
  so 7C3*4C4*4!=840
• 11 years agoHelpfull: Yes(3) No(3)
• answer is 840
• 11 years agoHelpfull: Yes(2) No(7)
• answer is 5!
• 11 years agoHelpfull: Yes(2) No(3)
• A1,A3,A5 as a group remaining can be arranged with group in 5!
  
  ans is 5!
• 11 years agoHelpfull: Yes(2) No(1)
• Total no of ways of arranging = 7! =5040
  Suppose the arrangement be (A5,A3,A1) which is contradictory to the given condition
  Taken the above arrangement as the whole group without changing their order.
  Now Req ways = Total ways - ways of arranging the above group
  = 7!-5!=4920
• 11 years agoHelpfull: Yes(0) No(5)