Question

A man has 7 relatives, 4 of them are ladies and 3 gentlemen; his wife has 7 relatives and 3 of them are ladies and 4 gentlemen. The number of ways they can invite a dinner party of 3 ladies and 3 gentlemen so that there are 3 of men's relatives and 3 of wife's relative is

• A. 455
• B. 565
• C. 485
• D. None of these

Answer 1

The correct option is C 485

There are 4 possibilities

i) 3 ladies from husband's side and 3 gentleman from wife's side .

No. of ways in this case${=}^4{C}_3{\times }^4{C}_3=4\times 4=16$.

ii) 3 gentleman from husband's side and 3 ladies from wife's side.

No. of ways in this case${=}^3{C}_3{\times }^3{C}_3=1\times 1=1$

iii) Two ladies & one gentleman from husband's side and one lady and two gentleman from wife's side.

No. of ways in this case$={(}^4{C}_2{\times }^3{C}_1\left){(}^3{C}_1{\times }^4{C}_2\right)=(3\times 6{)}^2=324$

iv) One lady and two gentlemen from husband's side and 2 ladies and one gentleman from wife's side.

No. of ways in this case$={(}^4{C}_1{\times }^3{C}_2)\times {(}^3{C}_2{\times }^4{C}_1)=(4\times 3{)}^2=144$

Total no. of ways are$=16+1+324+144=485$