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?

Answer 1

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:
$$10^6 - {6\choose1}9^6 + {6\choose2}9^6 - {6\choose4}9^6 + {6\choose5}9^6 = \boxed{625207}.$$