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) 484
(b) 485
(c) 468
(d) 469
Answers
Answered by
1
Answer:
(b) 485
Step-by-step explanation:
Required answer:
(3C3 x 4C0 x 3C3 x 4C0) + (3C3 x 4C1 x 3C3 x 4C1) + (3C1 x 4C2 x 3C1 x 4C2) + (3C0 x 4C3 x 3C0 x 4C3)
= 1 + 144 + 324 + 16
= 485
Similar questions