Final answer:
The number of ways six children can choose from three flavors of ice cream, with each flavor being selected by at least two children, is 48 different ways. This is calculated by assigning the flavor pairs in 3! ways and arranging the children within their flavor pairs in 2! ways per flavor.
Step-by-step explanation:
The question is a combinatorial problem where we are trying to count the number of ways six children can choose from three flavors of ice cream, given that each flavor must be selected by at least two children. To solve, we can see this as an application of partitioning problems or distributing problems. The partition we are looking for is of the form 2-2-2, which means two children selecting each of the three flavors. As the children are distinguishable but the scoops of the same flavor are not, we can first assign the flavors and then count the number of permutations of the children. There are 3! (six factorial) ways to assign the flavor pairs, and then within each flavor, the two children choosing the same flavor can be arranged in 2! ways.
Therefore, the total number of ways would be:
(3! ways to assign flavors) x (2! arrangements within first flavor) x (2! arrangements within second flavor) x (2! arrangements within third flavor).
Putting it all together:
3! (6 permutations of the flavor pairs) x 2! x 2! x 2! (8 permutations of the children within their flavor pairs)
Hence, the total number of ways is 3! x 2! x 2! x 2! = 6 x 2 x 2 x 2 = 48 different ways for the children to choose their ice cream flavors such that each flavor is chosen by at least two children.