# A train going from Delhi to Jaipur stops at 7 intermediate stations.Five persons enter the train during the journey with five different tickets of the same class.how many different sets of tickets they could had?Plz explain.

Swapnil Saxena
102 Points
11 years ago

According to questions, there r 7 intermediate stations, this means they can have 8 different kind of tickets.

Note: J,1,2,3,..7,D are the stations

D->1, D->2,D->3,D->4...D->7,D->J

Then if the passengers have the tickets of the same class can may be of same stations, no of tickets possible are 85

However if they have tickets of different stations of same coach the permutations are 8*7*6*5*4

Swapnil Saxena
102 Points
11 years ago

This is the condition if they al board the train from Delhi, However if they can board the train from any station . so here is the solution

Each person can have two choices 1) Boarding Station 2)Destination

if B.S= Delhi Des=1,2,3,4,5,6,7,Jaipur = 8 choices

if B.S.= 1  Des=2,3,4,5,6,7,Jaipur = 7 choices

if B.S.= 2  Des=3,4,5,6,7,Jaipur = 6 choices

...

thus the total type of tickets that a person can have =8+7+6+5+4+3+2+1 = 36choices

Similarly all the fove people can have these choices (May be recurring) .

Then the total choices may be 365

Howeverif they have different tickets then the permutation possible are 36*35*34*33*32

Menka Malguri
39 Points
11 years ago

The answer is (10C2 - 10),plz explain how?

290 Points
11 years ago

Hi Menka,

This is a very good Question.

May be you should try it on your own. You can get to know a lot of concepts through the thought process.

A small hint there.....

Say D, S1, S2,........, S7, J are the nine stations.

Then one set would be (just as an example) - D-S2, S3-S6, S2-S4, S1-S7, S5-S7.

Now from the 9 stations you have to find all such unique two station combination for the five persons.

Best Regards,

Mehul Trivedi
21 Points
7 years ago
D 1 2 3 4 5 6 7 J are stations.
Passenger entering at 1 can take 7 different destinations.
Passenger entering at 2 can take 6 different destinations.
Passenger entering at 3 can take 5 different destinations.
Passenger entering at 4 can take 4 different destinations.
Passenger entering at 5 can take 3 different destinations.
Passenger entering at 6 can take 2 different destinations.
Passenger entering at 7 can take 1 different destinations.
Total possible ways 7+6+5+4+3+2+1= 28 out of this total 5 passengers so 5 tickets. Answer 28C5.