Question

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 this party, is

(A)  468 

(B)  469 

(C)  484 

(D)  485 

Answer 1

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