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.485 B.468 C.469 D.484
ANSWER
Verified
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.
Complete step-by-step answer:
According to question, We have to see the number of possible ways in these above-mentioned cases; For case (1), When 2 ladies from X and 1 lady from Y, Then 1 man from X and 2 men from Y, then 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$ For case (3), When 1 lady from X and 2 ladies from Y, then 2 men from X and 1 man from Y, then 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$ Total number of possible ways= 10+324+144+1=485 Hence, option (A) is correct.
Note- For solving such questions, we have to make cases separately and then solve each case separately and then calculate by adding all the possible ways. We have to make sure that the cases do not get repeated otherwise it will lead to wrong answers. It is based on selection, not on arrangement.