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by sudhir3127 » Mon Jul 28, 2008 7:36 am
A train is going from Point A to point B and makes 5 stops on the way. If 3 persons enter the train during the journey with 3 different tickets.How many different set of tickets they may had?

ANswer after some discussion...
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by ricky » Mon Jul 28, 2008 8:05 am
Is the answer 56? :?
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by Chitts » Mon Jul 28, 2008 8:13 am
Sudhir I tried solving the sum below way..

Total possible tickets from A to B would be 6! = 720.
Now one has to choose 3 out of 6!(720) possibilities.
Hence answer is 720C3.

Is this approach correct?
Regards,
Chitts
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by pepeprepa » Mon Jul 28, 2008 1:19 pm
I would say 10.
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by TRANGZON » Mon Jul 28, 2008 2:39 pm
call 5 stop are a, b, c, d, e

counter possible tickets:
ab ac ad ae
bc bd be
cd ce
de

the answer is 4+3+2+1=10
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by sudhir3127 » Mon Jul 28, 2008 8:28 pm
Thanks a lot all for the discuss..
the answer is 15C3..

here it goes..

Ist station ... from where they could have bought tickets to any one of the five destination in 5 ways..

2nd station..
Ist station ... from where they could have bought tickets to any one of the four destination in 4 ways

sinilarly all.. hence it would be 5+4+3+2+1= 15 ways

hence 15 tickets are available out of which 3 tickets are to be choosen ..
therefore

15C3.

Please let me know if u guys are not claer on this...

Thanks All
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by pepeprepa » Mon Jul 28, 2008 9:45 pm
Ok I thought people entered at different stations but their final stop was B for all of them.
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