Andrea, Betty, Joyce, Karen, and Paula are starters on their school basketball team. How many different groups of three can be chosen for a newspaper photo?
You are looking for C(5,3)
which is 10
5! / 3!(5-3)! = 120 / 6(2) = 120/12 = 10
There are 10 different groups of three possible for the given combinations problem.
To find the number of different groups of three that can be chosen for a newspaper photo from the five starters on the basketball team, you can use the concept of combinations.
A combination is a selection of items from a larger set without regard to the order of the items. In this case, we are selecting three players from a group of five (Andrea, Betty, Joyce, Karen, and Paula).
The formula to calculate the number of combinations is:
nCr = n! / (r! * (n - r)!)
Where:
n is the total number of items in the set (5 in this case)
r is the number of items to be selected (3 in this case)
! denotes factorial, which is the product of an integer and all positive integers below it. For example, 5! = 5 * 4 * 3 * 2 * 1 = 120.
Plugging in the numbers, we have:
5C3 = 5! / (3! * (5 - 3)!)
= 5! / (3! * 2!)
= (5 * 4 * 3 * 2 * 1) / ((3 * 2 * 1) * (2 * 1))
= (120) / (6 * 2)
= 10
Therefore, there are 10 different groups of three that can be chosen for a newspaper photo from the five starters on the basketball team.