Question 1120228
.
<pre>
<U>Four men</U> PLUS <U>two men</U> are very special persons (at least at the context of this problem).


The rest  16 - (4+2) = 10 are not so special.


You can form the first team of 8 people by adding any 4 of the 10 people to these special four by  {{{C[10]^4}}} ways.


You can arrange then this team by  8! ways along one side of the table.


Now, you form the second team of 8 people by adding  6 = 10-4 of the remaining "not very special persons" to the two very special. 

In this case you just have no choice, so <U>you can form this team by ONLY ONE WAY</U>.


But next you can arrange this second team by  8!  ways along the second side of the table.


Finally, you can change "the sides of the table", which gives you an additional factor of 2.


Thus "the number of ways they can be seated" = {{{2*C[10]^4*8!*8!}}} = {{{2*((10*9*8*7)/(1*2*3*4))*8!*8!}}} = 2*210*8!*8!
</pre>