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10 speakers that includes 3 specific persons (A, B, C) are to speak in a function. Find the number of ways they can be arranged to speak if A has to speak before B and B has to speak before C. Second part: Find the number of ways these 10 speakers can be arranged if in addition to the above condition, B cannot speak immediately after A.

10 speakers that includes 3 specific persons (A, B, C) are to speak in a function. Find the number of ways they can be arranged to speak if A has to speak before B and B has to speak before C.
Second part: Find the number of ways these 10 speakers can be arranged if in addition to the above condition, B cannot speak immediately after A.

Grade:11

6 Answers

Harsh Patodia IIT Roorkee
askIITians Faculty 907 Points
8 years ago
For first case select three positions out of 10 and A,B,C can be arranged only in 1 way in those secletions i.e 10C1 *1. Then 7 remaining can be arranged in 7! ways.
For second one just subtract the cases from above by keeping Bimmediately after A
Natarajan
22 Points
8 years ago
Thanks   a ton for your excellent solution.For “The 7 remaining” the number of permutaions – I have a doubt here. Will it be 7! ways or will it be 10p7 ways? I feel it should be 10p7. Appreciate your clarification. 
Harsh Patodia IIT Roorkee
askIITians Faculty 907 Points
8 years ago
It will be 7! only 3positions have already been selected. So 7P7 i.e 7! ways.
Natarajan
22 Points
8 years ago
Just another thought of solving this pronlem
First let us take the 7 general speakers (leaving the 3 special speakers) and fill 7 positions out of 10 positions.. This will have 10p7 ways of filling. This way we make all 10 positions available for choosing for the general speakers. This leaves just 3! positions for the special speakers (A, B and C). However we can only choose the order A then B and then C. So it is only one way. So we have a total of 10p7 ways.
In 7! ways, since we have already selected 3 positions, I think we are denying the 3 positions choice for the general speakers.
mycroft holmes
272 Points
8 years ago
Your answer is right Natarajan. In Harsh’s approach, you choose 3 positions for the special three in  10C3 ways and then arrange the remaining 7 in 7! ways. So nr of ways = 7! X 10C3 which is equivalent to 10P7
Natarajan
22 Points
8 years ago
Holmes, Thanks for your confirmation. 
I think for the second part, We can find the no of ways ‘AB will occur together and remove them and that would be 7. I hope that is right?                                                                                                                                  

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