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. $485$
• B. $468$
• C. $469$
• D. $484$

Answer 1

The correct option is A $485$

The possible combinations are
$$\begin{array}{ccccc}X& 3L& 2L,1M& 1L,2M& 3M\\ Y& 3M& 2M,1L& 1M,2L& 3L\end{array}$$
Total number of ways $={}^4{C}_3\cdot {}^4{C}_3+{}^4{C}_2\cdot {}^3{C}_1\times {}^4{C}_2\cdot {}^3{C}_1+{}^4{C}_1\cdot {}^3{C}_2\times {}^4{C}_1\cdot {}^3{C}_2+{}^3{C}_3\cdot {}^3{C}_3$
$=485$