To me, without that, it is not a trivial problem! It will be another good problem...
Ok let us try again
This time I will try to solve your version of the problem. Tell me if I go wrong somewhere, ok?
Following your idea, first let us distribute each person only one fruit first. Since apples(and oranges) are identical, this amounts to choosing 6 people out of 10. The remaining get oranges(I have no choice for them. Choosing 6 people for apple fixes the orange guys!). So I have ways to do it. Now the remaining 2 fruits can go to anyone, this I can do in ways.
(I used the number of solutions to x1+x2+......+x10 = 2)
Totally
Now, Plato??
Okay, I thought this was much simpler than it was initally, I'm not that good at this combinatorial stuff, but here is my guess now:
Op1: Choose how many people get Apples initally: Either 4, 5, or 6, so 3 ways.
Op3: Give the rest oranges: 1 way
Op4: Give away remaining two pieces of fruit.
Case 1: 4 people get apples.
Op4.1.a: Pick the 4: ways
Op4.1.b: Now there are 2 more apples to give.
Case 1.1: 1 person gets both apples
Pick the person: 10 ways
Case 1.2: 2 people get apples
Pick the 2 people: ways
Case 2: 5 people get apples
Op4.2.a: Pick the 5: ways
Op4.2.b: Now there is 1 Orange, and 1 Apple to give.
Pick a person for Orange: 10 ways
Pick a person for Apple: 10 ways
Case 3: 6 people get apples
Op4.3.a: Pick the 6: ways
Now there are 2 more oranges to give.
Case 3.1: 1 person gets both oranges
Pick the person: 10 ways
Case 3.2: 2 people get oranges
Pick the 2 people: ways
Total:
3 ( 10C4 ( 10 + 10C2) + 10C5 ( 10^2 ) + 10C6 (10+10C2) ) = 144,900
Sorry, I meant to come back and expand my post.
Doing LOTS of different cases:
Case 1: Someone has 3 apples.
Any of the ten could have 3 apples, then split the other three apples among the remaining nine:
= 840 ways.
Case 2: Someone has 3 oranges.
Same as Case 1, 840 ways.
Case 3: Two people have two apples each.
Any two of the ten could have the two apples, then split the other two apples among the remaining eight:
= 1260 ways.
Case 4: Two people have two oranges each.
Same as Case 3, 1260 ways.
Case 5: Someone has two apples, and someone has one of each.
Any of the ten people could have two apples. Spread the six oranges among the other nine people, give the remaining people an apple each, then give the last apple to one of the six orange-holders:
= 5040 ways.
Case 6: Someone has two oranges, and someone has one of each.
Same as Case 5, 5040 ways.
Case 7: Two people have one of each.
Any two of the ten could have the mixed pair, then split the four apples and four oranges among the remaining eight people.
= 3150 ways.
Altogether, the seven cases gives 17430 total ways
(Oops, looks like I didn't add in Case 7 last time. That's what I get, huh?)
Good point.
Case 8: Someone has two apples and an orange.
Any of the ten people can have two apples and an orange, then split the other four apples among the other nine people:
= 1260 ways.
Case 9: Someone has two oranges and an apple.
Same as Case 8, 1260 ways.
Adding those in gives a total of 19950 ways.
I think that's all the possible cases.
Arrrrgh! Stop pointing out my mistakes!
Case 10: Someone has two apples, and someone has two oranges.
Any of the ten people could have two apples, and any of the remaining nine could have the two oranges. Then split the remaining four apples among the remaining eight people:
[tex]10*9*{8 \choose 4} = 6300 ways.
Altogether now, 26250 ways.
At some point, Plato, you'll be convinced I'm correct.
I think I did not count the identical setup that happens with oranges. I mean if I distribute the oranges first and then distribute apples, I will get your distribution. SO I need to double the solutions. However I think there will be double counting and I need to discount a few.... Am i on the right track?
Thanks