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k12345
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Antmavel
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k12345
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It sure is tough isnt it ??

Explanation:
At the first intermediate stop, tickets of next 6 stations will be available, at second stop tickets of next 5 stations will be available, and so on.
Therefore, total number of different tickets possible
= 6 + 5 + 4 + 3 + 2 + 1 = 21
5 persons could have had 21C5 different sets of tickets.
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k12345
It sure is tough isnt it ??

Explanation:
At the first intermediate stop, tickets of next 6 stations will be available, at second stop tickets of next 5 stations will be available, and so on.
Therefore, total number of different tickets possible
= 6 + 5 + 4 + 3 + 2 + 1 = 21
5 persons could have had 21C5 different sets of tickets.


Why do you consider it from first intermediate stop and not from the first stop (Look at Antmavels explanation)...That makes the problem crazy but this is not a tough one....
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I still think that the solution is C(10,5).

We have six intermediate stops. At each of these stops a number of people (from 0 to 5) can enter the train.

The problem is the same as a problem with six different marbles in a bag and we have to pick five with replacement. The order matters, so it is a combination with repetition problem.

I assume we do not have to consider the first stop as the stem says that "Five persons enter the train during this journey". If they enter during the journey, it cannot be the first stop.



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