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  1. #16
    GAMMA Mathematics
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    Quote Originally Posted by Plato View Post
    Now that is correct! But in your solution you assumed that there are 12 different pieces of fruit. If that were the case, it is a trivial problem of counting surjections.
    2^{10} meant each person had a possibility of having an apple or an organe. I never mentioned 12 different pieces of fruit in my last solution. There are 12 different people though.
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  2. #17
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    Quote Originally Posted by colby2152 View Post
    2^{10} meant each person had a possibility of having an apple or an orange.
    But it does not work that way. There are not 10 oranges.
    The 2^{10} counts as if there were ten of each.
    But there are only six of each.
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  3. #18
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    Quote Originally Posted by Plato View Post
    But it does not work that way. There are not 10 oranges.
    The 2^{10} counts as if there were ten of each.
    But there are only six of each.
    Good point, we should probably go with the fact that you can have either four and six, five and five, or six and four apples and oranges respectively for the set of ten people.
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  4. #19
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    Quote Originally Posted by Plato View Post
    in how many ways can 6 apples and 6 oranges be distributed among 10 persons provided each get any one of those?
    Without that is a trivial problem.
    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 {10 \choose 6} ways to do it. Now the remaining 2 fruits can go to anyone, this I can do in {11 \choose 9} = 55 ways.
    (I used the number of solutions to x1+x2+......+x10 = 2)
    Totally 55 \cdot {10 \choose 6}

    Now, Plato??
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  5. #20
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    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: 10\choose 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: 10\choose 2 ways



    Case 2: 5 people get apples
    Op4.2.a: Pick the 5: 10\choose 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: 10\choose 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: 10\choose 2 ways


    Total:
    3 ( 10C4 ( 10 + 10C2) + 10C5 ( 10^2 ) + 10C6 (10+10C2) ) = 144,900
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  6. #21
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    Quote Originally Posted by Isomorphism View Post
    Totally 55 \cdot {10 \choose 6}
    Now, Plato??
    Well that is getting somewhere. But you realize that it is possible for one of the ten to receive three apples. Can that happen in your setup?
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  7. #22
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  8. #23
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    Quote Originally Posted by Henderson View Post
    14,280
    What is that? A guess?
    Explain yourself, please.
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  9. #24
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    Quote Originally Posted by Plato View Post
    What is that? A guess?
    Explain yourself, please.
    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:
    10*{9 \choose 3} = 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:
    {10 \choose 2}{8 \choose 2} = 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:
    10*{9 \choose 6}*6 = 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.
    {10 \choose 2}{8 \choose 4} = 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?)
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  10. #25
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    You are correct that there are ten ways a person can receive three pieces of fruit.
    But do you realize that it could be any one of four cases: three apples, three oranges, two apples & one orange or two oranges & one apple?
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  11. #26
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    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:
    10*{9 \choose 4} = 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.
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  12. #27
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    Have you thought about how many cases there are for two people to get two each?
    (aa,aa);(oo,oo);(aa,oo);(ao,oa);etc.
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  13. #28
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    Quote Originally Posted by Plato View Post
    Have you thought about how many cases there are for two people to get two each?
    (aa,aa);(oo,oo);(aa,oo);(ao,oa);etc.
    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.
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  14. #29
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    Quote Originally Posted by Plato View Post
    Well that is getting somewhere. But you realize that it is possible for one of the ten to receive three apples. Can that happen in your setup?
    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
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  15. #30
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    Quote Originally Posted by Henderson View Post
    Altogether now, 26250 ways.
    That is the total I got.
    I have been looking for a way to do this with a generating function.
    There must be one, but I cannot find a suitable model.
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