In a lottery game, a player picks 9 numbers from 1 to 43. How many different choices does the player have if order doesn't matter?
43C9 = ?
To calculate the number of different choices the player has, we need to use combinations.
In this case, the player is selecting 9 numbers from a set of 43 numbers, and since the order doesn't matter, it is a combination problem.
The formula for combinations is:
C(n, k) = n! / (k!(n-k)!)
Where n is the total number of items in the set and k is the number of items to be selected.
In this case, n = 43 (the total number of numbers to choose from) and k = 9 (the number of numbers the player selects).
Using the formula, we can calculate the number of different choices:
C(43, 9) = 43! / (9!(43-9)!)
Simplifying this expression gives:
C(43, 9) = (43 * 42 * 41 * 40 * 39 * 38 * 37 * 36 * 35) / (9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1)
After canceling out common factors, we can calculate the result:
C(43, 9) = 186,884,815
Therefore, the player has 186,884,815 different choices when selecting 9 numbers from 1 to 43 in a lottery game where the order doesn't matter.