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?
+195 We can solve this problem using the Principle of Inclusion-Exclusion (PIE). We first find the total number of ways each child can choose a flavor, which is 3 choices for each child, or 3^6 = 729 total possible combinations.
Next, we find the number of ways that no flavor is selected by exactly three children. This means that either no flavor is selected by any of the children, or one flavor is selected by two children and the other two flavors are selected by one child each.
Case 1: No flavor is selected by any of the children
There is only one way this can happen - each child chooses a different flavor. There are 3 choices for the first child, 2 choices for the second child (since one flavor has already been chosen), and so on, for a total of 3! = 6 possible combinations.
Case 2: One flavor is selected by two children and the other two flavors are selected by one child each
There are 3 ways to choose which flavor is selected by two children. Then there are 5 choices for the first child (any flavor except the one chosen by the two children), 4 choices for the second child (any remaining flavor except the one chosen by the two children), and so on, for a total of 3 * 5 * 4 * 3 * 2 * 1 = 360 possible combinations.
Therefore, the number of ways that no flavor is selected by exactly three children is 6 + 360 = 366.
Finally, we use PIE to find the number of ways that at least one flavor is selected by exactly three children. This is given by:
Total - (No flavor is selected by exactly three children)
= 729 - 366
= 363
Therefore, there are 363 ways that each child can choose a flavor for their scoop of ice cream so that some flavor of ice cream is selected by exactly three children.
