Six children are each offered a single scoop of any of 3 flavors of ice cream from the Combinations Creamery. How many ways can each child choose a flavor for their scoop of ice cream so that some flavor of ice cream is selected by exactly three children?
Let's use the three ice cream flavors A, B, and C. We want exactly three children to choose the same flavor. This means that the other three children must choose different flavors, otherwise, there would be more than three children with the same flavor.
We have three cases to consider:
1. Three children choose flavor A, and the other three children choose one scoop each of flavors B and C.
2. Three children choose flavor B, and the other three children choose one scoop each of flavors A and C.
3. Three children choose flavor C, and the other three children choose one scoop each of flavors A and B.
For each case, we need to determine the number of ways to assign the ice cream flavors to the six children.
Case 1:
- Choose 3 children out of 6 to receive flavor A: C(6, 3) = 6! / (3! * (6-3)!) = 20 ways.
- The other three children must receive one scoop each of flavors B and C. There are 2 ways to arrange this (BCB or CBC).
So there are 20 * 2 = 40 arrangements for Case 1.
Case 2:
- Choose 3 children out of 6 to receive flavor B: C(6, 3) = 20 ways.
- The other three children must receive one scoop each of flavors A and C. There are 2 ways to arrange this (ACA or CAC).
So there are 20 * 2 = 40 arrangements for Case 2.
Case 3:
- Choose 3 children out of 6 to receive flavor C: C(6, 3) = 20 ways.
- The other three children must receive one scoop each of flavors A and B. There are 2 ways to arrange this (ABA or BAB).
So there are 20 * 2 = 40 arrangements for Case 3.
Now, we add up the arrangements from all three cases:
Total arrangements = 40 + 40 + 40 = 120
Therefore, there are 120 ways for the six children to choose ice cream flavors so that exactly one flavor is selected by three children.
