Q. 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 :
Solution:
$X(4\, L\, 3\, G)$
$Y(3\, L\, 4\, G)$
$3\, L\, 0\, G$
$0\, L\, 3\, G$
$2\, L\, 1\, G$
$1\, L\, 2\, G$
$1\, L\, 2\, G$
$2\, L\, 1\, G$
$0\, L\, 3\, G$
$3\, L\, 0\, G$
Required number of ways
$={ }^{4} C_{3} \cdot{ }^{4} C_{3}+\left({ }^{4} C_{2} \cdot{ }^{3} C_{1}\right)^{2}+\left({ }^{4} C_{1} \cdot{ }^{3} C_{2}\right)^{2}+\left({ }^{3} C_{3}\right)^{2}$
$=16 + 324 + 144 + 1$
$= 485$
| $X(4\, L\, 3\, G)$ | $Y(3\, L\, 4\, G)$ |
| $3\, L\, 0\, G$ | $0\, L\, 3\, G$ |
| $2\, L\, 1\, G$ | $1\, L\, 2\, G$ |
| $1\, L\, 2\, G$ | $2\, L\, 1\, G$ |
| $0\, L\, 3\, G$ | $3\, L\, 0\, G$ |