+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
+50 From https://brainly.com/question/30733513
Using combination, the number of ways of placing 8 counters in the square is 12 ways and the number of ways of placing 12 counters in the square is 8 ways
What is the number of ways of placing 8 counters in the square
a) To place 8 counters in the 4 x 4 grid, with exactly two counters in each row, we can first choose the two columns in the first row that will have counters. Using combination, there are {4 C 2} = 6 ways to do this. Then, we choose the two columns in the second row that will have counters, but we need to make sure that these columns are not the same as the ones chosen in the first row. There are only two columns left to choose from, so there are only two ways to do this. Similarly, we choose the two columns in the third row that will have counters, making sure that they are different from the columns chosen in the first two rows. There is only one column left to choose from, so there is only one way to do this. Finally, the columns for the fourth row are uniquely determined. Therefore, the total number of ways to place 8 counters in the grid with exactly two counters in each row is:
6 * 2 = 12
(b) To place 12 counters in the 4 x 4 grid, with exactly three counters in each column, we can start by choosing the three rows that will have counters in the first column. Using combination, there are {4 C3} = 4 ways to do this. Then, we choose the three rows that will have counters in the second column, but we need to make sure that these rows are not the same as the ones chosen for the first column. There are only two rows left to choose from, so there are only two ways to do this. Similarly, we choose the three rows that will have counters in the third column, making sure that they are different from the rows chosen for the first two columns. There is only one row left to choose from, so there is only one way to do this. Finally, the rows for the fourth column are uniquely determined. Therefore, the total number of ways to place 12 counters in the grid with exactly three counters in each column is:
4 * 2 = 8
Thus, there are 12 ways to place 8 counters in the grid with exactly two counters in each row, and 8 ways to place 12 counters in the grid with exactly three counters in each column.
