## Task B-2.7.

On a shelf, there are six different chocolates and three different boxes of sweets. Each of the three children can choose for dessert either two chocolates or one box of sweets. In how many ways can the three children choose their desserts?
## Solution.

When choosing, we consider the following cases.

First case: each of the three children can choose two chocolates.

In this case, the first child can choose the first chocolate in 6 ways, the second in 5 ways, which totals $6 \cdot 5$ ways. However, each pair of chocolates has been counted twice because, for example, ab and ba are the same choice of chocolates for one child. Therefore, the total number of ways the first child can choose two chocolates is $\frac{6 \cdot 5}{2}=15$.

The second child has 4 chocolates available and can choose two chocolates in $\frac{4 \cdot 3}{2}=6$ ways. Since only two chocolates are left, the third child can choose two chocolates in 1 way.

The total number of choices of two chocolates each is $15 \cdot 6 \cdot 1=90$ because for each of the 15 ways the first child can choose two chocolates, there are 6 ways the second child can choose two chocolates.

Second case: two children can choose two chocolates each, and one child can choose one box of chocolates.

In three ways, we can first choose the child who will choose one box of chocolates. That child can choose the box in 3 ways. Next, one of the remaining two children can choose two chocolates in $\frac{6 \cdot 5}{2}=15$ ways, and the other in $\frac{4 \cdot 3}{2}=6$ ways. Therefore, the total number of choices in this case is $3 \cdot 15 \cdot 6 \cdot 3=810$.

Third case: two children can choose one box of chocolates each, and one child can choose two chocolates.

Similarly to the previous case, we will first choose the child who will choose two chocolates in 3 ways, and that child can do so in $\frac{6 \cdot 5}{2}=15$ ways. Next, one of the remaining two children can choose one box of chocolates in 3 ways, and the other in 2 ways.

Therefore, the total number of choices in this case is $3 \cdot 15 \cdot 3 \cdot 2=270$.

Fourth case: each of the three children can choose one box of chocolates.

Obviously, the number of choices in this case is $3 \cdot 2 \cdot 1=6$.

1 point

Finally, the total number of ways the children can choose dessert is obtained by summing the total number of choices from the four cases mentioned, which is the number

$$
90+810+270+6=1176
$$

## COUNTY COMPETITION IN MATHEMATICS

3rd grade - high school - B variant

## February 26, 2024.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE TASK, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH IN AN APPROPRIATE MANNER.