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 of $X$ and $Y$ are in this party, is

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

Answer 1

The correct option is D $485$

	$Ladies$	$Men$
$X$	$4$	$3$
$Y$	$3$	$4$

Case(I):When $3$ ladies from $X$ and $3$ men from $Y$

No. of ways ${=}^4{C_3}^4{C}_3=16$

Case (ii): when $2$ ladies from $X$ and $1$ lady from $Y$ , then $1$ man from $X$ and $2$ men from $Y$.

No. of ways ${=}^4{C_2}^3{C_1}^4{C}_2=324$

Case (iii) when $1$ lady from $X$ and $2$ ladies from $y$, then from $X$ and $1$ man from $Y$

No. of ways ${=}^4{C_1}^3{C_2}^3{C_2}^4{C}_1=144$

Case (iv):when $0$ lady from $X$ i.e $3$ men from $X$ and $3$ ladies from $Y$

No. of ways ${=}^3{C_3}^3{C}_3=1$

$\therefore$ Total no. of ways $=10+324+144+1$

$=485$

Hence, option $D$ is correct answer.