A train going from station A to station B has 10 stations in between as halts. Nine persons enter the train duning the journey with nine different tickets of the same class. How many sorts of tickets they may have had.

Answer

To solve the problem of how many different sorts of tickets the nine persons may have had, we can break down the solution step by step.

### Step 1: Understand the Problem
We have a train journey from station A to station B, with 10 stations in between. Each of the 9 persons has a different ticket, and we need to determine how many different types of tickets they could have.

### Step 2: Identify Available Tickets
At each station (S1 to S10), the number of available tickets decreases as the journey progresses:
- At S1, there are 10 possible destinations (S2 to S10, plus B).
- At S2, there are 9 possible destinations (S3 to S10, plus B).
- At S3, there are 8 possible destinations (S4 to S10, plus B).
- This pattern continues until S10, where there is only 1 destination (B).

### Step 3: Calculate Total Tickets
We can calculate the total number of different tickets available by summing the available tickets at each station:
- Total tickets = 10 (at S1) + 9 (at S2) + 8 (at S3) + 7 (at S4) + 6 (at S5) + 5 (at S6) + 4 (at S7) + 3 (at S8) + 2 (at S9) + 1 (at S10)

This can be simplified using the formula for the sum of the first n natural numbers:
\[ \text{Sum} = \frac{n(n + 1)}{2} \]
where \( n = 10 \):
\[ \text{Sum} = \frac{10(10 + 1)}{2} = \frac{10 \times 11}{2} = 55 \]

### Step 4: Choose 9 Different Tickets
Now that we know there are 55 different tickets available, we need to choose 9 different tickets from these 55. The number of ways to choose 9 tickets from 55 is given by the combination formula:
\[ \binom{n}{r} = \frac{n!}{r!(n - r)!} \]
where \( n = 55 \) and \( r = 9 \):
\[ \text{Number of ways} = \binom{55}{9} \]

### Final Answer
Thus, the total number of different sorts of tickets the nine persons may have had is:
\[ \binom{55}{9} \]