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 :

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

Tardigrade Admin · Accepted Answer

X(4 L 3 G) Y(3 L 4 G) 3 L 0 G 0 L 3 G 2 L 1 G 1 L 2 G 1 L 2 G 2 L 1 G 0 L 3 G 3 L 0 G Required number of ways = 4 C3 ⋅ 4 C3+( 4 C2 ⋅ 3 C1)2+( 4 C1 ⋅ 3 C2)2+( 3 C3)2 =16 + 324 + 144 + 1 = 485

$X (4 L 3 G)$	$Y (3 L 4 G)$
$3 L 0 G$	$0 L 3 G$
$2 L 1 G$	$1 L 2 G$
$1 L 2 G$	$2 L 1 G$
$0 L 3 G$	$3 L 0 G$

468

469

484

485

X(4L3G) Y(3L4G) 3L0G 0L3G 2L1G 1L2G 1L2G 2L1G 0L3G 3L0G Required number of ways =4C3​⋅4C3​+(4C2​⋅3C1​)2+(4C1​⋅3C2​)2+(3C3​)2 =16+324+144+1 =485

$X (4 L 3 G)$ $Y (3 L 4 G)$

$3 L 0 G$ $0 L 3 G$

$2 L 1 G$ $1 L 2 G$

$1 L 2 G$ $2 L 1 G$
$0 L 3 G$ $3 L 0 G$

Required number of ways $=^{4} C_{3} \cdot^{4} C_{3} + (^{4} C_{2} \cdot^{3} C_{1})^{2} + (^{4} C_{1} \cdot^{3} C_{2})^{2} + (^{3} C_{3})^{2}$
$= 16 + 324 + 144 + 1$
$= 485$