Problem 7.6.1. In a row, there are 10 boxes. These boxes contain balls of two colors: red and blue. In some boxes, all the balls may be of the same color; there are no empty boxes. It is known that in each subsequent box (from left to right), there are no fewer balls than in the previous one. It is also known that there are no two boxes with the same set of red and blue balls. How many blue and how many red balls are in the rightmost box, if the total number of red balls in all the boxes is 11 and the total number of blue balls is 13?
Answer: 1 red ball, 3 blue balls.

Solution. Among all the boxes, there can be:

- a maximum of 2 boxes with one ball (one with a red ball, the other with a blue ball),
- a maximum of 3 boxes with two balls (one with two red balls, another with one red and one blue, and the third with two blue balls),
- a maximum of 4 boxes with three balls (one with three red balls, another with two red and one blue, the third with one red and two blue, and the fourth with three blue balls).

Thus, in the first nine boxes, there will be at least $2 \cdot 1 + 3 \cdot 2 + 4 \cdot 3 = 20$ balls. Then, in the rightmost box, there will be no more than $11 + 13 - 20 = 4$ balls. However, from the above reasoning, it follows that if a box contains 1, 2, or 3 balls, it must be one of the first nine boxes. Therefore, the rightmost box must contain exactly 4 balls, and the first nine boxes must contain exactly 20 balls.

It is not difficult to understand that the first nine boxes contain 10 red and 10 blue balls. Then, in the rightmost box, there will be $11 - 10 = 1$ red and $13 - 10 = 3$ blue balls. $\square$