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 friend. Number of ways in which X & 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. 485
• B. 468
• C. 469
• D. 484

Answer 1

The correct option is A 485

Given, X has 7 friends, 4 of them are ladies and 3 are men while Y has 7 friends, 3 of them are ladies and 4 are men.
$\therefore$ Total number of required ways
${}^3{C}_3{\times }^4{C}_0{\times }^4{C}_0{\times }^3{C}_3{+}^3{C}_2{\times }^4{C}_1{\times }^4{C}_1{\times }^3{C}_2{+}^3{C}_1{\times }^4{C}_2{\times }^4{C}_2{\times }^3{C}_1{+}^3{C}_0{\times }^4{C}_3{\times }^4{C}_3{\times }^3{C}_0=1+144+324+16=485$