nice and easy problem, but it took me 3 hours to solve it!!! suppose that you have $n$ countries and $m$ committees, namely $A_1,...,A_m$ satisfying the condition, where $m$ is an odd number. now add another country with three members $\{1,2,3\}$, and consider these committees: $B_1=A_1\cup \{1\},...,B_m=A_m\cup \{1\},B_{m+1}=A_1\cup \{2\},...,B_{2m}=A_m\cup \{2\}$. these committees satisfy the condition, so we have now $2m$ committees from $n+1$ countries. also, if $k$ is an even integer that $k\le m$, then, as above, these committes make a cycle of length $2(k-1)$ with $n+1$ countries: $B_1=A_1 \cup \{3\}, B_2=A_2\cup \{1\}, B_3=A_3\cup \{2\},...,B_{k-2}=A_{k-2}\cup \{1\}, B_{k-1}=A_{k-1}\cup \{2\}, B_k=A_k\cup \{3\}, B_{k+1}=A_{k-1}\cup \{1\},...,B_{2(k-1)}=A_2\cup \{2\}$. now starting from the first, we have $1$ country and $3$ committees, next, we have $2$ countries and $6$ committees, next $3$ countries and $10$ committees,... and at last $11$ countries and $1990$ committees.

can you explain more about how you'll arrive at $ 11 $ countries and $ 1990 $ committees please?

goodar2006 wrote: nice and easy problem, but it took me 3 hours to solve it!!! suppose that you have $n$ countries and $m$ committees, namely $A_1,...,A_m$ satisfying the condition, where $m$ is an odd number. now add another country with three members $\{1,2,3\}$, and consider these committees: $B_1=A_1\cup \{1\}, B_2=A_2\cup\{2\}, B_3=A_3\cup\{1\}, ...,B_m=A_m\cup \{1\},B_{m+1}=A_1\cup \{2\},...,B_{2m}=A_m\cup \{2\}$. these committees satisfy the condition, so we have now $2m$ committees from $n+1$ countries. also, if $k$ is an even integer that $k \leq m$, then, as above, these committes make a cycle of length $2(k-1)$ with $n+1$ countries: $B_1=A_1 \cup \{3\}, B_{k} = A_{k} \cup\{3\}$, arrange the rest of ${B_i}$ from $1<i<=2k-2, i \neq k$ as $A_2, A_3, ... A_{k-1}, A_{1}, ... A_{k-2}$ Alternate the appendages (from the n+1th country) by alternating between 1 and 2 in this set. Now starting from the first, we have $1$ country and $3$ committees, next, we have $2$ countries and $6$ committees, next $3$ countries and $10$ committees,... and at last $11$ countries and $1990$ committees. The recursion actually generates a cycle of length 2390 for n = 11, which should apparently also imply the existence of a cycle of length 1990 since 2390 > 1990. However, I don't quite see why the existence of a larger cycle should imply the existence of a smaller one. Also I fixed the indices (the one's Goodar wrote for odd m were just plain wrong, I'm sure he mistyped or something) so I wrote them to match my indices.

I misread the question

Wizard_32 wrote: A nice problem! The answer is yes. We prove the result for all $n \geq 6$. Firstly, let the countries be $C_1, C_2, C_3, \cdots C_{k}$, and the represntatives be $a_{i1}, a_{i2}, a_{i3} \in C_i$ for all $1 \le i \le n$. Common sense says that if a member is common to two committees, then the person belongs to the same country (because they are the same person). Hence we can classify all the members according to their countries, and countries in our figure are columns. Hence we only need to check common members for the respective columns (i.e. countries).

, which can be easily handled. \begin{tabular}{|c|c|c|c|c|c|c|} \hline & $C_1$ & $C_2$ & $C_3$ & $C_4$ & $C_5$ & $C_6$\\ \hline $A_1$ & $\alpha_{11}$ & $\alpha_{21}$ & $\alpha_{31}$ & $\alpha_{42}$ & $\alpha_{51}$ & $\alpha_{63}$ \\ \hline $A_2$ & $\alpha_{12}$ & $\alpha_{22}$ & $\alpha_{32}$ & $\alpha_{43}$ & $\alpha_{52}$ & $\alpha_{62}$ \\ \hline $A_3$ & $\alpha_{11}$ & $\alpha_{23}$ & $\alpha_{33}$ & $\alpha_{41}$ & $\alpha_{53}$ & $\alpha_{61}$ \\ \hline $A_4$ & $\alpha_{12}$ & $\alpha_{21}$ & $\alpha_{31}$ & $\alpha_{42}$ & $\alpha_{52}$ & $\alpha_{62}$ \\ \hline $A_5$ & $\alpha_{11}$ & $\alpha_{22}$ & $\alpha_{32}$ & $\alpha_{43}$ & $\alpha_{51}$ & $\alpha_{63}$ \\ \hline $A_6$ & $\alpha_{12}$ & $\alpha_{23}$ & $\alpha_{31}$ & $\alpha_{41}$ & $\alpha_{52}$ & $\alpha_{62}$ \\ \hline $A_7$ & $\alpha_{11}$ & $\alpha_{21}$ & $\alpha_{32}$ & $\alpha_{42}$ & $\alpha_{53}$ & $\alpha_{63}$ \\ \hline $A_8$ & $\alpha_{12}$ & $\alpha_{22}$ & $\alpha_{31}$ & $\alpha_{41}$ & $\alpha_{51}$ & $\alpha_{62}$ \\ \hline $A_9$ & $\alpha_{11}$ & $\alpha_{23}$ & $\alpha_{32}$ & $\alpha_{42}$ & $\alpha_{53}$ & $\alpha_{61}$ \\ \hline $A_{10}$ & $\alpha_{12}$ & $\alpha_{21}$ & $\alpha_{31}$ & $\alpha_{41}$ & $\alpha_{51}$ & $\alpha_{63}$ \\ \hline $A_{11}$ & $\alpha_{13}$ & $\alpha_{22}$ & $\alpha_{32}$ & $\alpha_{43}$ & $\alpha_{52}$ & $\alpha_{61}$ \\ \hline \end{tabular} P.S. The $6th$ column is probably unnecessary, but I added it just because in my construction with $5$ columns, the $3rd$ and $11th$ rows were disjoint. That's why the other members of the $6th$ column were arbitrarily chosen with only the third condition in mind. You proved it is possible to make $11$ commities using $6$ country.The problem is something else... .And I don't think You can generalize this method to work for the problem(The bound on problem is very sharper than this ...)The only way to solve this is inductivly I think Since it is tooooo hard to think for a direct construction.

Taha1381 wrote: You proved it is possible to make $11$ commities using $6$ country.The problem is something else... .And I don't think You can generalize this method to work for the problem(The bound on problem is very sharper than this ...)The only way to solve this is inductivly I think Since it is tooooo hard to think for a direct construction. Note that in my construction, the important conditions are all covered: (i) each committee has $ n$ members, one from each country; (ii) no two committees have the same membership; (iv) if $ 1 < |i - j| < m - 1,$ then committees $ A(i)$ and $ A(j)$ have at least one member in common These three conditions are satisfied for the first 11 countries themselves since any two non-adjacent countries have at least one common element. Now, we can add the countries after $C_{11}$. without having to worry about the condition (iv) Also, each pair of countries have at least one different element, so no matter how we add on the following countries, no two countries are exactly the same. Now, we can add the countries after $C_{11}$. without having to worry about the condition (ii) Also, the condition 3 is not violated yet, and will not be violated further if we add the next columns as: (bascially alternate the members) \begin{tabular}{|c|c|c|c|c|c|c|} \hline & $C_i$ \\ \hline $A_1$ & $\alpha_{i1}$ \\ \hline $A_2$ & $\alpha_{i2}$ \\ \hline $A_3$ & $\alpha_{i1}$ \\ \hline $A_4$ & $\alpha_{i2}$ \\ \hline $A_5$ & $\alpha_{i1}$ \\ \hline $A_6$ & $\alpha_{i2}$ \\ \hline $A_7$ & $\alpha_{i1}$ \\ \hline $A_8$ & $\alpha_{i2}$ \\ \hline $A_9$ & $\alpha_{i1}$ \\ \hline $A_{10}$ & $\alpha_{i2}$ \\ \hline $A_{11}$ & $\alpha_{i3}$ \\ \hline \end{tabular} 800th post

Wizard_32 wrote: Taha1381 wrote: You proved it is possible to make $11$ commities using $6$ country.The problem is something else... .And I don't think You can generalize this method to work for the problem(The bound on problem is very sharper than this ...)The only way to solve this is inductivly I think Since it is tooooo hard to think for a direct construction. Note that in my construction, the important conditions are all covered: (i) each committee has $ n$ members, one from each country; (ii) no two committees have the same membership; (iv) if $ 1 < |i - j| < m - 1,$ then committees $ A(i)$ and $ A(j)$ have at least one member in common These three conditions are satisfied for the first 11 countries themselves since any two non-adjacent countries have at least one common element. Now, we can add the countries after $C_{11}$. without having to worry about the condition (iv) Also, each pair of countries have at least one different element, so no matter how we add on the following countries, no two countries are exactly the same. Now, we can add the countries after $C_{11}$. without having to worry about the condition (ii) Also, the condition 3 is not violated yet, and will not be violated further if we add the next columns as: (bascially alternate the members) \begin{tabular}{|c|c|c|c|c|c|c|} \hline & $C_i$ \\ \hline $A_1$ & $\alpha_{i1}$ \\ \hline $A_2$ & $\alpha_{i2}$ \\ \hline $A_3$ & $\alpha_{i1}$ \\ \hline $A_4$ & $\alpha_{i2}$ \\ \hline $A_5$ & $\alpha_{i1}$ \\ \hline $A_6$ & $\alpha_{i2}$ \\ \hline $A_7$ & $\alpha_{i1}$ \\ \hline $A_8$ & $\alpha_{i2}$ \\ \hline $A_9$ & $\alpha_{i1}$ \\ \hline $A_{10}$ & $\alpha_{i2}$ \\ \hline $A_{11}$ & $\alpha_{i3}$ \\ \hline \end{tabular} 800th post You are adding committees not countries we have $11$ countries but a lot more (1990) committees.

O god seems like I misread the question Thanks for pointing out my mistake!

Solution: Yes, it is possible to have a cycle of $1990$ committees with $11$ $\bullet \ $ Construction: Suppose $A_1,A_2, \dots , A_{m}$ is an $m-$cycle with $k$ countries. We present an algorithm which shows that it is possible to construct a $2m-2$-cycle involving exactly $k+1$ countries. Denote by $\{C_1,C_2,C_3\}$ the three members of the $(k+1)^{\text{th}}$ country. Then, observe that the committees: $$\{A_1,C_3\} , \{A_2,C_2\},\{A_3,C_1\},\{A_4,C_2\},\{A_5,C_1\}, \dots ,\{A_{m},C_3\},\dots, \{A_3,C_2\},\{A_2,C_1\}$$also satisfy all constraints of the problem. Hence we have successfully created a $2m-2$-cycle with $(k+1)$ countries. Claim: A $6$-cycle is possible with $2$ countries. Proof. Let $\{B_1,B_2,B_3\}$ and $\{C_1,C_2,C_3\}$ represent the members of the two countries respectively. Then consider the committees: $ A_1 = \{B_1 , C_1\} ; A_2 = \{B_2 , C_2\} ; A_3 = \{B_1, C_3\} ; A_4 =\{ B_2 , C_1\} ; A_5 =\{ B_1 , C_2\} ;A_6 =\{ B_2, C_3\} $ It can be verified with ease that all these work. Thus, with $2$ countries we can create a $6$-cycle and the recursions gives that with $11$ countries we can create a $2050$-cycle. But since we desire a $1990$-cycle, we must remove some of the committees. Again, we present an algorithm which shows that we can remove,in each step , $2$ committees without worrying about the constraints. $\rightarrow$ From the cycle, $\{A_1,C_3\} , \{A_2,C_2\},\{A_3,C_1\},\{A_4,C_2\},\{A_5,C_1\}, \dots ,\{A_{m},C_3\},\dots, \{A_3,C_2\},\{A_2,C_1\}$ we remove the first and the last committees and replace the $C_{j}$ of the second committee from the left with $C_3$. i.e. The new cycle is $\{A_2,C_3\} , \{A_3,C_1\} , \dots , \{A_3,C_2\}$. Since $2050-1990 =60 $ is even, we are done $\ \blacksquare$

i hv a not so elegant solution, but you can just calculate the linearity of expectation and it just exceeds one