Correct option (D) 485
The situation is depicted as in the following figure:
Case I: 3L from X side and 3M from Y side. Therefore,
4C3 x 4C3 = 4 x 4 = 16.
Case II: 3M from X side 3L from Y side. Therefore,
3C3 x 3C3 = 1 x 1 = 1
Case III: 2L and 1M from X side and 2M and 1L from Y side. Therefore,
(4C2 x 3C1) x (4C2 x 3C1) = (6 x 3) x (6 x 3) = 18 x 18 = 324
Case IV: 2M and 1L from X side and 1M and 2L from Y side.
(3C2 x 4C1) x (4C1 x 3C2) = (3 x 4) x (4 x 3)= 12 x 12 = 144
Therefore, the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party, is
Case I + Case II + Case III + Case IV
= 16 + 1 + 324 + 144 = 485