10-1. Piglet has balloons of five colors. He managed to arrange them in a row in such a way that for any two different colors in the row, there will always be two adjacent balloons of these colors. What is the minimum number of balloons Piglet could have?
Answer: 11 balls.

Solution. Consider the balls of color $a$. Their neighbors must be balls of all 4 other colors. But one ball can have no more than two neighbors, so there must be at least 2 balls of color $a$. This is true for each of the 5 colors, so there must be at least 10 balls in total.

Notice that some ball of color $b$ is at the end of the row. It has only one neighbor, so two balls of this color have no more than 3 neighbors. Therefore, to meet the conditions, we need to add at least one more ball of color $b$. Thus, we need at least 11 balls in total.

We will show that 11 balls are sufficient. Let $a, b, c, d, e$ be the colors of the balls. They can be arranged, for example, as follows: $a b c d e c a d b e a$.

Criteria. Only the answer - 0 points. Estimation (that there are no fewer than 11 balls) - 4 points. Example of the required arrangement - 3 points.