Question

A train going from Vijayawada to Hyderabad stops at nine intermediate stations. Six people enter the train during the journey with six different tickets of the same class. The number of different tickets they may have

1. $^{11}{C_6}$
2. $^{45}{C_6}$
3. $^9{C_6}$
4. $^{10}{C_6}$

Answer 1

We will find the number of different tickets that a person can buy when he enters any station. Also, we have to consider the possibilities of all the stations as people can enter from any station. We will add all the possibilities to find the number of tickets that a person can buy at all 9 intermediate stations. Then, use a combination to determine the number of 6 different tickets possible.

Complete step-by-step answer:
We are given that there are 9 intermediate sessions between Vijayawada to Hyderabad.
A person can enter at any of the nine intermediate stations.
When a person enters station 1, he can buy tickets for any one stop from 2 to 9 and to Hyderabad, thus, he has 9 choices.
Similarly, there will be 8 choices when the person enters the second station and so on.
The person entering from station 9 will have only 1 choice, that is to last stop, Hyderabad.
We will find the total number of different tickets possible when a person enters at any station.
The number of different tickets at all the stations can be $9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45$
But, we have to select 6 different tickets from 45 tickets.
Hence, we will use combination to find the number of different tickets the passengers can have
$^{45}{C_6}$
Hence, option B is correct.

Note: Concept of combination when the order of the arrangement does not matter. Also, if we have the expression $^n{C_r}$, then it is equivalent to $\dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, where $n! = n.\left( {n - 1} \right).\left( {n - 2} \right).....3.2.1$. Here, we cannot use permutation because we do not want tickets in any fixed order. Also, do not forget to consider the last stop, Hyderabad while calculating the number of different tickets.