What is the probability A speaks before B and B before C?

[gaganpreetsingh]

Three persons A,B,C are to speak at the function along with 5 other persons. If the persons speak in random order, what is the probability that A speaks before B and B speaks before C?

I am unable to begin with calculating the favourable possibilities.

 

[EnumaElish]

There are 8 persons, call them A, B, C, D, E, F, G, H.

In how many different orderings can you put these 8 people?

How many of these orderings satisfy order(A) < order(B) < order(C)?

 

[HallsofIvy]

Since all you are concerned with is the position of A, B, C, I don't think you need to worry about the other 5. How many orders for 3 people, A, B, C, are there? How many of those have precisely ABC in that order?

 

[EnumaElish]

But, isn't ABDEFGHC a different outcome than ABDEFGCH? By your logic, these two would've counted as one.

 

[0rthodontist]

Take the spots that A, B, and C talk in as all the same kind of spot--say X. Then find all the ways to place 3 X's out of 8 spots, and then find how many total ways the other speakers can be arranged once the 3 X's are placed.

Edit: though actually, this will lead to the same answer as what Ivy and Integral said to do.

 

[HallsofIvy]

But, isn't ABDEFGHC a different outcome than ABDEFGCH? By your logic, these two would've counted as one.

If you consider all 8 people, then, yes, they are different outcomes- and the numerator of the probability fraction will be different than if you only consider A, B, C. But then you would have to consider ACDEFGHB (where they are not in ABC order) as different from ACDEFGBH. The denominator of the fraction is also different- the result is the same.

To take a simple example: suppose there are 4 people, A, B, C, D speaking. What is the probability that A, B, C speak in the order ABC?

There are, of course 4!= 24 different ways to order 4 things:
ABCD ABDC ACBD ACDB ADBC ADCB
BACD BADC BCAD BCDA BDAC BDCA
CBAD CBDA CABD CADB CDBA CDAB
DBCA DBAC DCBA DCAB DABC DACB

Of those, ABCD, ABDA, ADBC, DABC have A, B, C in that order: the probability that A, B, C will speak in that order is 4/24= 1/6.

If instead we count only A, B, C, we find 3!= 6 different orders and only 1 (ABC) has them in that order: Again 1/6.

Given any n people speaking in random order, the probability that a given 3 will speak in a given order is 1!/3!= 1/6. If you use all orders for n people as the denominator and all orders for n people that have the same 3 people in the given order, the numerator and denominator are both multiplied by n!/3!.