A man has $7$ relatives, $4$ women and $3$ men. His wife also has $7$ relatives, $3$ women and $4$ men. The number of ways in which they can invite $3$ men and $3$ women so that they both invite three each is

485

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584

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720

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1024

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Solution

The correct option is B $485$
Let the husband have the following relatives
$4$ women $W_{1}, W_{2}, W_{3}, W_{4}$ and $3$ men $M_{1}, M_{2}, M_{3}$

Let the wife have the following relatives
$3$ women $w_{1}, w_{2}, w_{3}$ and $4$ men $m_{1}, m_{2}, m_{3}, m_{4}$

Favorable cases of $3$ men and $3$ women-
$(3 W, 3 m) + (1 M, 2 W, 2 m, 1 w) + (2 M, 1 W, 1 m, 2 w) + (3 M, 3 w)$

$^{4} C_{3}^{4} C_{3} +^{3} C_{1}^{4} C_{2}^{4} C_{2}^{3} C_{1} +^{3} C_{2}^{4} C_{1}^{4} C_{1}^{3} C_{2} +^{3} C_{3}^{3} C_{3} = 485$

Hence, option A.

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