A man `X` has `7` friends, `4` of them are ladies and `3` are men. His wife `Y` also has `7` friends, `3` of them are ladies and 4 are men. Assume `X` and `Y` have no common friends. Then 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 the party, is : 469 (2) 484 (3) 485 (4) 468

Explanation

There are `4` possible cases.

Case - 1 : When `X` invites all three men and `Y` invites all three ladies.

Total number of ways in this case `= C(3,3)*C(3,3) = 1*1 = 1`

Case - 2 : When `X` invites two men and one lady and `Y` invites one men and two ladies.

Total number of ways in this case `= C(3,2)*C(4,1)*C(4,1)*C(3,2) = 3*4*4*3 = 144`

Case - 3 : When `X` invites one man and two ladies and `Y` invites one lady and two men.

Total number of ways in this case `= C(3,1)*C(4,2)*C(3,1)*C(4,2) = 3*6*3*6 = 324`

Case - 4 : When `X` invites all three ladies and `Y` invites all three men.

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