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 and 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:

• A. 484
• B. 485
• C. 468
• D. 469

Answer 1

The correct option is B 485

$X\to 4w+3M$

$Y\to 3w+4M$

total in party $\Rightarrow 3M+3W$ and $3$ of $X$ and $3$ of $Y$

$X$ $Y$

$3M$ $3W$ $\to {}^3{C}_3.{}^3{C}_3=1$

$2M1W$ $1M2W\to {}^3{C}_2.{}^4{C}_1.{}^4{C}_1.{}^3{C}_2=14$

$1M2W$ $2M1W\to {}^3{C}_1.{}^4{C}_2.{}^4{C}_2.{}^3{C}_1=3$

$3W$ $3M$ $\to {}^4{C}_3.{}^4{C}_3=16$

$\therefore$

total number of ways is

$1+144+324+16$

$=485$

$\therefore$ Option $B$ is correct