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 two children?
+12 We can use the complement principle to solve this problem. That is, we can first count the number of ways that no flavor of ice cream is selected by exactly two children, and then subtract that from the total number of possible choices.
There are three flavors of ice cream, and each child can choose any of the three flavors. So the total number of possible choices is 3^6.
Now, let's count the number of ways that no flavor of ice cream is selected by exactly two children. There are two cases to consider:
Case 1: No flavor of ice cream is selected by more than one child. In this case, the first child can choose any of the three flavors. The second child can choose any of the two remaining flavors. The third child can choose any of the two remaining flavors. The fourth child can choose any of the two remaining flavors. The fifth child can choose any of the two remaining flavors. The sixth child can choose the last remaining flavor. So there are 32222*1 = 48 ways to make choices in this case.
Case 2: One flavor of ice cream is selected by exactly two children, and the other two flavors are each selected by one child. In this case, there are three ways to choose which flavor will be selected by exactly two children. Once that is chosen, there are 4 ways to choose which two children will select that flavor. Then, the remaining two flavors can be assigned to the remaining four children in 22 = 4 ways. So there are 34*4 = 48 ways to make choices in this case as well.
Therefore, the total number of ways to make choices so that some flavor of ice cream is selected by exactly two children is 3^6 - 48 - 48 = 585. Answer: \boxed{585}. MI Bridges App
