Let both the sides be named $A\:and \:B$

Having seated the 4 men on side A and 2 on B, we are left with 10 persons.

We can choose 4 of them for side A in $^{10}C_4$ ways. and remaining 8

can be seated automatically on side B.

After selecting the persons for each side, arrangement is done.

On each side arrangement is done is $8!$ ways.

$\therefore$ The required no. of seating arrangements = $^{10}C_4.8!8!$