Six children are each offered a single scoop of any of flavors of ice cream from the Combinatorial Creamery. In 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?
We can solve this problem using complementary counting. First, we find the total number of ways that the children can choose their ice cream flavors, which is $10^6$ since each child has 10 choices.
Next, we count the number of ways that no flavor is selected by exactly three children. Suppose that each flavor is selected by one, two, four, five, or six children. There are ${6\choose1}$ ways to choose which flavor is selected by one child, ${6\choose2}$ ways to choose which two flavors are selected by two children, ${6\choose4}$ ways to choose which four flavors are selected by four children, and ${6\choose5}$ ways to choose which five flavors are selected by five children. For each of these cases, there are $9^6$ ways for the children to choose their flavors, since each child can choose from 9 flavors (since one flavor is already chosen by someone else).
Therefore, the number of ways that no flavor is selected by exactly three children is
$${6\choose1}9^6 - {6\choose2}9^6 + {6\choose4}9^6 - {6\choose5}9^6.$$
Finally, we subtract this from the total to get the number of ways that some flavor is selected by exactly three children:
$$10^6 - {6\choose1}9^6 + {6\choose2}9^6 - {6\choose4}9^6 + {6\choose5}9^6 = \boxed{625207}.$$
To find the number of ways the children can choose a flavor of ice cream such that some flavor is selected by exactly three children, we need to consider the possible scenarios.
Scenario 1: Three children choose one flavor, and the other three children choose different flavors.
In this scenario, there are 6 flavors to choose from for the three children, which can be selected in 6 choose 1 ways. The remaining 3 children can choose their flavors from the remaining 5 flavors (excluding the one chosen by the first three) in 5 choose 3 ways. Therefore, the total number of ways for this scenario is 6 choose 1 * 5 choose 3.
Scenario 2: Three children choose one flavor, and the other three children choose another flavor.
In this scenario, there are 6 flavors to choose from for the first three children, which can be selected in 6 choose 1 ways. The remaining 3 children can choose their flavors from the remaining 5 flavors (excluding the one chosen by the first three) in 5 choose 3 ways. Therefore, the total number of ways for this scenario is 6 choose 1 * 5 choose 3.
Since these two scenarios are mutually exclusive (they cannot happen at the same time), we can sum up the total number of ways as follows:
Total number of ways = Scenario 1 + Scenario 2
= (6 choose 1 * 5 choose 3) + (6 choose 1 * 5 choose 3)
= 6C1 * 5C3 + 6C1 * 5C3
= (6 * 10) + (6 * 10)
= 60 + 60
= 120
Therefore, there are 120 ways for the children to choose their flavors of ice cream such that some flavor is selected by exactly three children.