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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 comman 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 :
$484$
$485$
$468$
$469$
Solution
$X \to 4$ ladied $Y \to 3$
$X \to 3$ men $Y \to 4$
Possible cases for $X$ are
$(1)$ $3$ ladies, $0$ man
$(2)$ $2$ ladies, $1$ man
$(3)$ $1$ lady, $2$ men
$(4)$ $0$ ladies, $3$ men
Possible cases for $Y$ are
$(1)$ $0$ ladies, $3$ men
$(2)$ $1$ ladies, $2$ men
$(3)$ $2$ lady, $1$ man
$(4)$ $3$ ladies, $0$ man
No. of ways ${ = ^4}{C_3}{ \cdot ^4}{C_3} + {{(^4}{C_2}{ \cdot ^3}{C_1})^2} + {{(^4}{C_1}{ \cdot ^3}{C_2})^2} + {{(^3}{C_3})^2}$
$ = 16 + 324 + 144 + 1 = 485$
