How many ways can you give out 4 oranges, 2 bananas and 2 lemons so that everyone get 2 fruits and nobody gets 2 oranges **How many ways can you give out 4 oranges, 2 bananas and 2 lemons so that everyone get 2 fruits and no one gets 2 oranges** My solution: Ok so if no one can have 2 oranges, everyone gets 1 in every case. 1) OB OB OL OL 2) OL OL OB OB 3) OB OL OB OL 4) OL OB OL OB 5) OL OB OB OL 6) OB OL OL OB **= 6 ways** I wonder if someone knows a more "mathematically" approach to solve this problem.

From your problem, you know that every person has to get an orange and one other fruit. Then your problem reduces to "In how many ways can we give two bananas to 4 people?" In this case is equivalent to "In how many ways can we give two lemons to 4 people?"The answer is $$\binom 42=6$$