9.2. In a kindergarten, each child was given three cards, each of which had either "MA" or "NYA" written on it. It turned out that 20 children could form the word "MAMA" from their cards, 30 children could form the word "NYANYA," and 40 children could form the word "MANYA." How many children had all three cards the same?
Answer: 10 children.

Solution. Let's denote the number of children who received three "MA" cards as $x$, two "MA" cards and one "NA" card as $y$, two "NA" cards and one "MA" card as $z$, and three "NA" cards as $t$. Then, the word "MAMA" can be formed by all children from the first and second groups and only them, the word "NANA" - by all children from the third and fourth groups and only them, and the word "MANA" - by all children from the second and third groups and only them. Therefore, $x+y=20, z+t=30, y+z=40$, so the desired number is $x+t=(x+y)+(z+t)-(y+z)=20+30-40=10$ children.

Grading criteria. A guessed answer with some examples is not to be evaluated. Formulating but not solving the system of equations: 3 points. Misinterpreting the problem: 0 points.