2. (CAN 1) Given $n$ countries with three representatives each, $m$ committees $A(1), A(2), \ldots A(m)$ are called a cycle if
(i) each committee has $n$ members, one from each country;
(ii) no two committees have the same membership;
(iii) for $\dot{i}=1,2, \ldots, m$, committee $A(i)$ and committee $A(i+1)$ have no member in common, where $A(m+1)$ denotes $A(1)$ :
(iv) if $1<|i-j|<m-1$, then committees $A(i)$ and $A(j)$ have at least one member in common.
Is it possible to have a cycle of 1990 committees with 11 countries?
2. We will call a cycle with $m$ committees and $n$ countries an $(m, n)$ cycle. We will number the delegates from each country with numbers $1,2,3$ and denote committees by arrays of these integers (of length $n$ ) defining which of the delegates from each country is in the committee. We will first devise methods of constructing larger cycles out of smaller cycles.
Let $A_{1}, \ldots, A_{m}$ be an $(m, n)$ cycle, where $m$ is odd. Then the following is a $(2 m, n+1)$ cycle:
$$
\left(A_{1}, 1\right),\left(A_{2}, 2\right), \ldots,\left(A_{m}, 1\right),\left(A_{1}, 2\right),\left(A_{2}, 1\right), \ldots,\left(A_{m}, 2\right) .
$$

Also, let $A_{1}, \ldots, A_{m}$ be an $(m, n)$ cycle and $k \leq m$ an even integer. Then the cycle
$$
\begin{array}{c}
\left(A_{1}, 3\right),\left(A_{2}, 1\right),\left(A_{3}, 2\right), \ldots,\left(A_{k-2}, 1\right),\left(A_{k-1}, 2\right), \\
\left(A_{k}, 3\right),\left(A_{k-1}, 1\right),\left(A_{k-2}, 2\right), \ldots,\left(A_{2}, 2\right)
\end{array}
$$
is a $(2(k-1), n+1)$ cycle.
Starting from the ((1),(2),(3)) cycle with parameters $(3,1)$ we can sequentially construct larger cycles using the shown methods. The obtained cycles have parameters as follows:
$$
(6,2),(10,3), \ldots,\left(2^{k}+2, k\right), \ldots,(1026,10),(1990,11) .
$$

Thus there exists a cycle of 1990 committees with 11 countries.