In this lottery game, the player is choosing 6 numbers from a total of 43, and the order in which they pick the numbers does not matter. This is a combination problem, where we calculate the number of ways to choose $k$ items from $n$ items without regard to order, using the combination formula:

$C(n, k) = \frac{n!}{k!(n-k)!}$

For this problem:

• $n = 43$ (total numbers to choose from)
• $k = 6$ (numbers to be chosen)

Plugging in the values:

$C(43, 6) = \frac{43!}{6!(43-6)!} = \frac{43!}{6! \times 37!}$

Now, let's calculate that:

$C(43, 6) = \frac{43 \times 42 \times 41 \times 40 \times 39 \times 38}{6 \times 5 \times 4 \times 3 \times 2 \times 1}$

Calculating the numerator and the denominator:

$C(43, 6) = \frac{4,082,940}{720} = 5,671$

Therefore, the player has 5,671 different choices.

Would you like to explore this further or have any other questions? Here are some related topics:

1. How does the probability change if the order of numbers matters?
2. What if the player picks 5 numbers instead of 6?
3. How do the combinations change if the total numbers increase to 49?
4. How do combinations differ from permutations?
5. How does changing one parameter (like choosing 7 numbers) affect the total choices?

Tip: In combination problems, remember that order doesn't matter. Always use combinations when dealing with order-independent scenarios.