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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:
- 484
- 485
- 468
- 469
Correct answer: 485
Solution
X gives 3 friends, Y gives 3 friends; need 3 ladies + 3 men total. If X gives k ladies it gives (3-k) men, forcing Y to give (3-k) ladies and k men. Sum_{k=0..3} C(4,k)C(3,3-k)*C(3,3-k)C(4,k) = 1+144+324+16 = 485.
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