Hungarian tram and bus tickets have 9 possible locations for holes.

Passengers need to validate their tickets on their own using a punching machine that creates holes on the ticket. Transportation officials randomly travel around town and ask for the passengers' validated tickets. The tickets do not expire. In theory, the ticket needs to be inserted into the punching machine with the red arrow on top. In practice, this does not matter since the officials do not care about the direction. So, inserting the ticket with the red arrow on the bottom creates the same ticket. A fee evader wants to collect every possible validated ticket and use the appropriate one every time he/she travels. How many different validated tickets are needed if every punching machine in town creates 2 holes on a ticket?

The fee evader needs to collect how many
 tickets.

We are tasked with determining how many distinct validated tickets a fee evader needs to collect if every punching machine in town creates 2 holes on a ticket, and the orientation of the ticket does not matter (i.e., inserting it upside down is treated as the same ticket).

### Step 1: Total possible combinations of 2 holes in 9 positions
Since there are 9 possible locations for holes on the ticket, and each machine creates exactly 2 holes, we first calculate the number of ways to choose 2 positions from the 9 possible ones. This is a combination problem, and the number of combinations is given by the formula for combinations:

\[
\binom{9}{2} = \frac{9 \times 8}{2} = 36
\]

So, if orientation mattered, there would be 36 different ways to punch 2 holes on the ticket.

### Step 2: Adjusting for indistinguishable orientations
However, the problem states that the ticket’s orientation does not matter, meaning that a ticket punched in one way is considered the same as one punched in its flipped version. We need to account for this by identifying and removing the symmetries caused by flipping the ticket.

For every pair of punched positions, there is another ticket where the holes appear in the opposite positions (due to flipping). Therefore, we need to divide by 2 to eliminate these duplicate configurations. This is valid because, for each distinct pair of holes, flipping the ticket produces an equivalent result.

\[
\text{Distinct tickets} = \frac{36}{2} = 18
\]

### Conclusion:
The fee evader needs to collect 18 distinct validated tickets to cover all possible combinations of punched holes, considering that the orientation of the ticket does not matter.

Would you like more details or have any questions? Here are 5 related questions that expand on this problem:

1. How would the solution change if the machine punched 3 holes instead of 2?
2. What if the ticket had 12 possible positions for holes instead of 9?
3. How many distinct tickets would there be if the orientation did matter?
4. How would you calculate the number of distinct tickets if the machine punched 1 hole instead of 2?
5. What if some holes overlapped due to multiple validations—how would that affect the solution?

**Tip:** When solving combination problems with symmetries, always check whether orientation or other properties affect the distinctiveness of the outcomes.

Hungarian tram and bus tickets have 9 possible locations for holes. Passengers need to validate their tickets on their own using a punching machine that creates holes on the ticket. Transportation officials randomly travel around town and ask for the passengers' validated tickets. The tickets do not expire. In theory, the ticket needs to be inserted into the punching machine with the red arrow on top. In practice, this does not matter since the officials do not care about the direction. So, inserting the ticket with the red arrow on the bottom creates the same ticket. A fee evader wants to collect every possible validated ticket and use the appropriate one every time he/she travels. How many different validated tickets are needed if every punching machine in town creates 2 holes on a ticket?

Explore how to solve a combinatorics problem involving 9 possible hole locations on tram tickets where a punching machine creates 2 holes. Learn how to calculate the number of distinct tickets needed, considering ticket orientation symmetry. Suitable for students in Grades 10-12.

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