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 X and Y are in this party is.A.485B.468C.469D.484

Hint: Here, in this case we have to form the possible cases to get the solution. The possible cases for can be:

1.When 2 ladies from X and 1 lady from Y, Then 1 man from X and 2 men from Y.
2.When 3 ladies from X and 3 men from Y.
3.When 1 lady from X and 2 ladies from Y, then 2 men from X and 1 man from Y.
4.When 0 ladies from X i.e. 3 men from X and 3 ladies from Y.

Number of possible ways$= $^4{C_2}^3{C_1}^3{C_1}^4{C_2} = 324 For case (2), When 3 ladies from X and 3 men from Y, then Number of possible ways =$^4{C_3}^4{C_3} = 16$
Number of possible ways$= $^4{C_1}^3{C_2}^3{C_2}^4{C_1} = 144 For case (4), When 0 lady from X i.e. 3 men from X and 3 ladies from Y, then Number of possible ways =$^3{C_3}^3{C_3} = 1$