11.2. Several married couples came to the New Year's Eve party, each of whom had from 1 to 10 children. Santa Claus chose one child, one mother, and one father from three different families and took them for a ride in his sleigh. It turned out that he had exactly 3630 ways to choose the required trio of people. How many children could there be in total at this party?

(S. Volchonkov)
Answer: 33.

Solution: Let there be $p$ married couples and $d$ children at the party (from the condition, $d \leqslant 10 p$). Then each child was part of $(p-1)(p-2)$ trios: a mother could be chosen from one of the $p-1$ married couples, and with a fixed choice of mother, a father could be chosen from one of the $p-2$ remaining couples. Therefore, the total number of trios is $d \cdot(p-1)(p-2)=3630$. Since $d \leqslant 10 p$, we get $3630 \leqslant 10 p^{3}$, that is, $p^{3} \geqslant 363>7^{3}$. Thus, $p \geqslant 8$.

Next, the number $3630=2 \cdot 3 \cdot 5 \cdot 11^{2}$ has two divisors $p-1$ and $p-2$, differing by 1. If one of these divisors is divisible by 11, then the other gives a remainder of 1 or 10 when divided by 11. Then it is coprime with 11, and thus divides $2 \cdot 3 \cdot 5=30$ and is at least 10. It is easy to see that this divisor can only be 10; then $p-2=10, p-1=11$ and $d=3630 / 110=33$.

If, however, neither of the numbers $p-2$ and $p-1$ is divisible by 11, then the number $2 \cdot 3 \cdot 5=30$ is divisible by their product; this contradicts the fact that $p \geqslant 8$.

Comment: The equality $d(p-1)(p-2)=3630$ was justified - 4 points.