Question

A max $X$ 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

Answer: (A)

$485$

$X$ has $4$ ladies friends and $3$ men friends.

$Y$ has $3$ ladies friends and $4$ men friends.

$X$ and $Y$ have to invite total $3$ ladies and $3$ men to the party and each of $X$ and $Y$ should bring in $3$ people to the party. Consider the cases as $(L_X,M_X,L_Y,M_Y)$ where $L$ and $M$ denote the lady or man corresponding to $X$ and $Y$. The possible cases are as follows:

Cases	No. of ways	Numerical Value
$(3,0,0,3)$	${}^4{C}_3{}^4{C}_3$	$16$
$(2,1,1,2)$	${}^4{C}_2{}^3{C}_1{}^3{C}_1{}^4{C}_2$	$324$
$(1,2,2,1)$	${}^4{C}_1{}^3{C}_2{}^3{C}_2{}^4{C}_1$	$144$
$(0,3,3,0)$	${}^3{C}_3{}^3{C}_3$	$1$

Hence, the total number of ways $=16+324+144+1=485$.