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, we need to determine how many different types of tickets the 9 persons can have when boarding the train at various stations between station A and station B.

### Step-by-step Solution:

1. **Identify the Stations**:
The train travels from station A to station B with 10 stations in between. Therefore, we have a total of 12 stations (A, 10 intermediate stations, and B).

2. **Determine the Ticket Options**:
Each person can board the train at any of the 10 intermediate stations. The tickets are valid for travel from the boarding station to any station further down the line (including the final destination, station B).

3. **Calculate the Number of Ticket Options for Each Person**:
- The first person can board at any of the 10 stations.
- The second person can board at any of the remaining 9 stations.
- The third person can board at any of the remaining 8 stations.
- This pattern continues until the ninth person, who can board at the last remaining station.

4. **Sum the Options**:
The total number of options for the tickets can be calculated as:
- Total options = 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 55.
This is the sum of the first 10 natural numbers.

5. **Choose 9 Tickets from the Total Options**:
Since there are 9 different tickets, we need to find the number of ways to choose 9 tickets from the 55 available options. This can be calculated using the combination formula:
\[ \text{Number of ways} = \binom{55}{9} \]

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