The problem asks us to determine how many different ways Carlota can use her 8 plays on exactly 5 different games, given that the order of plays matters. Here's how we can approach this:

Step 1: Choosing the 5 different games

Out of the 10 available games, Carlota needs to choose 5 distinct games. The number of ways to choose 5 games from 10 is given by the combination formula: $\binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$ So, there are 252 ways to select 5 different games.

Step 2: Distributing 8 plays among the 5 games

Next, we need to distribute 8 plays among the 5 selected games, with the condition that each game is played at least once. This is an example of a stars and bars problem with the restriction that each game gets at least one play.

To handle this, we first assign 1 play to each of the 5 games. This uses up 5 plays, leaving 3 remaining plays to distribute among the 5 games. This is now a standard stars and bars problem, where we need to distribute 3 indistinguishable plays among 5 games. The number of ways to do this is given by the formula for combinations with repetition: $\binom{3 + 5 - 1}{5 - 1} = \binom{7}{4} = \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} = 35$ So, there are 35 ways to distribute the remaining 3 plays among the 5 games.

Step 3: Arranging the plays in order

Since the order of the plays matters, we must consider the number of ways to arrange the 8 plays. Once we have assigned the number of plays to each game, the number of ways to arrange these plays is the number of ways to arrange 8 objects where some are identical (since some games might be played multiple times).

If the plays are divided as, say, $x_1$ plays for the first game, $x_2$ for the second game, etc., the number of ways to arrange these plays is the multinomial coefficient: $\frac{8!}{x_1! x_2! x_3! x_4! x_5!}$ However, we don't need to compute this explicitly for every possible distribution, because we already handled the arrangement of plays implicitly by considering the distinct distributions and the stars and bars method in Step 2.

Step 4: Combining the results

To find the total number of ways to use Carlota's 8 plays, we multiply the number of ways to:

1. Choose 5 games,
2. Distribute the 8 plays among the 5 games (ensuring each game is played at least once).

Thus, the total number of ways is: $\binom{10}{5} \times \binom{7}{4} = 252 \times 35 = 8,820$

Final Answer:

Carlota can use her 8 plays in 8,820 different ways.

Would you like further details or have any questions about the solution? Here are some additional questions related to the problem:

1. How would the solution change if the order of the plays didn't matter?
2. What if Carlota had only 7 plays instead of 8 but still needed to use exactly 5 games?
3. How would the solution change if Carlota could play fewer than 5 games but up to 5?
4. How would the solution change if Carlota could play more than 8 times?
5. What if the arcade had 12 games instead of 10?

Tip: The stars and bars method is very useful when distributing indistinguishable objects (such as extra plays) into distinguishable categories (like the games).