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A man X has 7 friends, 4 of them are lad...

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 the party, is : 469 (2) 484 (3) 485 (4) 468

A

485

B

468

C

469

D

484

Text Solution

Verified by Experts

The correct Answer is:
A
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) xx ""^(4)C_(0) xx ""^(4)C_(0) xx ""^(3)C_(3) + ""^(3)C_(2) xx ""^(4)C_(1) xx ""^(4)C_(1) xx ""^(3)C_(2)`
`+ ""^(3)C_(1) xx ""^(4)C_(2) xx ""^(4)C_(2) xx ""^(3)C_(1) + ""^(3)C_(0) xx ""^(4)C_(3) xx ""^(4)C_(3) xx ""^(3)C_(0)`
`=1 + 144 + 324 + 16 = 485`
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