Since the player picks five numbers from 0 to 9 and the order doesn’t matter, we can approach this as a combination problem with repetition (since a number can be chosen more than once).

We need to find the number of ways to choose 5 numbers from 10 possible choices (0 through 9), with repetition allowed and without considering the order.

The formula for combinations with repetition is given by:

$\binom{n + r - 1}{r}$

where:

• $n$ is the number of options (in this case, 10),
• $r$ is the number of choices made (in this case, 5).

Substituting the values:

$\binom{10 + 5 - 1}{5} = \binom{14}{5}$

Calculating $\binom{14}{5}$:

$\binom{14}{5} = \frac{14!}{5!(14 - 5)!} = \frac{14!}{5! \times 9!}$

Breaking this down:

$\frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2 \times 1} = 2002$

Thus, the player has 2002 different choices when picking five numbers from 0 to 9 if order does not matter.

Would you like further clarification or details on this approach?

Here are some related questions you might find useful:

1. What would be the number of choices if the player picked 4 numbers instead of 5?
2. How would the solution change if order did matter?
3. If numbers could not repeat, how many different choices would the player have?
4. What if the player picked numbers from 0 to 15 instead?
5. How does this calculation relate to combinations without repetition?

Tip: For problems involving combinations with repetition, remember to adjust the formula to account for the possibility of repeated selections, using $\binom{n + r - 1}{r}$.