$6$ Countries participate a world tournament .
Each country has $4$ players. One Cricket player , One Rugby player , one Volleyball player and one Football player.
Need to select a team of $8$ members.
Rule :
First four members must be selected from two countries out of 6 countries and 2 from each selected two countries.
Then we must select last four members from the rest of countries , by selecting two countries and 2 from each selected countries.
My attempt
Number of ways selecting two countries out of $6$ countries $$=\binom{6}{2}$$
Number of combinations for selecting two players from each selected country $$=\binom{4}{2}\binom{4}{2}$$
Now there are four countries left.
So number of ways selecting two countries from the left $$=\binom{4}{2}$$
Number of combinations for selecting two players from each selected country $$=\binom{4}{2}\binom{4}{2}$$
So altogether all combinations $$=\binom{6}{2}\binom{4}{2}\binom{4}{2}\cdot \binom{4}{2}\binom{4}{2}\binom{4}{2}$$
But one of my friends told me that the his answer is $$=\binom{6}{4}\cdot \binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{2}$$
What is correct ? This is confusing. Need a help.
If just $2$ members each from $4$ countries are to be selected, your friend is correct, but
the rules are very specific, and imply that there is a hierarchy in the choices (reason not specified).
A plausible assumption could be that we are to have two labelled groups, $A$ and $B$,
and your computation is correct according to the rules.
Your friend chooses all $4$ participating countries at one go,
but they would need to divide these $4$ into two labelled groups in $\binom42$ ways