3. The mathematical committee consists of $2 n$ members, $n \geqslant 3$. It is known that each member of the committee is in a dispute with exactly one other member of the committee (this relationship is symmetric). In how many ways can the committee be divided into three subcommittees: one for creating problems, one for evaluating problems, and one for organizing the competition, such that each subcommittee has at least two members and no two members of the committee who are in a dispute are in the same subcommittee?
3. Let's first find the number of ways to select the desired committees if we ignore the requirement that each committee must have at least two members. Each pair of mathematicians who are in a dispute can be arranged into three committees in a total of $3 \cdot 2=6$ ways (we place the first mathematician in any committee, and for the second, we choose one of the remaining two), so the total number of ways is $6^{n}$. From this number, we need to subtract the selections where the number of members in some committee is 0 or 1. Note, first of all, that there can be at most one such committee (if two committees were such, in the third we would have at least $2 n-2$ members of the commission, and since for $n \geqslant 3$ we have $2 n-2>n$, some of the members in that committee would be in a dispute with each other). The number of selections where some committee has no members is $3 \cdot 2^{n}$ (we first fix which of the 3 committees will have no members, and then each pair of mathematicians in a dispute can be arranged in the remaining two committees in 2 ways). The number of selections where some committee has only one member is $2 n \cdot 3 \cdot 2 \cdot 2^{n-1}=3 n \cdot 2^{n+1}$ (the lone mathematician can be chosen in $2 n$ ways, placed in one of 3 committees, the person they are in a dispute with in any of the remaining 2 committees, and then the remaining $2 n-2$ mathematicians can be placed in $2^{n-1}$ ways). It follows that the solution to the problem is the number $6^{n}-3 \cdot 2^{n}-3 n \cdot 2^{n+1}=$ $6^{n}-3 \cdot 2^{n}(2 n+1)$.