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QJEEM17Ma9e23: JEE Main 2017 Mathematics - Class 11 JEEM17Ma9e23 Solution: A man X has 7 friends, 4 of them are ladies and 3 are men. His…
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:
Multiple Choice Options
- A 468
- B 469
- C 484
- D 485
Step-by-Step Verified Solution
| X(7 Friends) | Y(7 Friends) | |||
|---|---|---|---|---|
| 4 Ladies | 3 Men | 3 Ladies | 4 Men | |
| Case 1 | 3 | 0 | 0 | 3 |
| Case 2 | 0 | 3 | 3 | 0 |
| Case 3 | 2 | 1 | 1 | 2 |
| Case 4 | 1 | 2 | 2 | 1 |
In Case 1, Case 2, Case 3 and Case 4, total 6 friends are present and 3 from X and 3 from Y and among those 6 friend 3 are ladies and 3 are men in every case.
= 16 + 1 + 324 + 144 = 485