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Math Help - counting rule?

  1. #1
    Senior Member sfspitfire23's Avatar
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    counting rule?

    There are 3 hangers and 5 shirts in a room. In how many ways, can these 5 shirts be arranged among the three hangers?

    Would we solve this with the counting rule where we take 5(3)=15?


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  2. #2
    Senior Member yeKciM's Avatar
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    there's 10 combinations

     \displaystyle \frac {5!}{3!\cdot2!}=10
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  3. #3
    MHF Contributor Unknown008's Avatar
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    On the first hanger, you can put one from any of the five shirts. This gives 5 ways.
    On the second hanger, you can put one from any of the four remaining shirts. This gives 4 more ways.
    On the third hanger, you can put one from any of the three remaining shirts. This gives 3 more ways.

    In total, you have 5x4x3 = 60 ways.
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  4. #4
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    There are 3 hangers and 5 shirts in a room. In how many ways, can these 5 shirts be arranged among the three hangers?

    Would we solve this with the counting rule where we take 5(3)=15?


    I don't think it's clear what assumptions we're supposed to make. Supposing we can put multiple shirts on a hanger, a hanger can be empty, we must use all shirts, hangers are identical, and shirts are identical, there are 5 ways.

    {0,0,5}
    {0,1,4}
    {0,2,3}
    {1,1,3}
    {1,2,2}

    If the shirts are non-identical (which is a natural assumption), it becomes 41 ways

    {0,0,5} --- 1
    {0,1,4} --- 5
    {0,2,3} --- C(5,2)
    {1,1,3} --- C(5,3)
    {1,2,2} --- 5*C(4,2)/2

    and there are many variations of assumptions.
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  5. #5
    Senior Member yeKciM's Avatar
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    hmmmmm...
    how can it bee so many ways when there is only 10 ways to to put them on 3 hangers ... if position of shirt on hanger doesn't have any influence
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  6. #6
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by yeKciM View Post
    there's 10 combinations

     \displaystyle \frac {5!}{3!\cdot2!}=10
    Quote Originally Posted by yeKciM View Post
    hmmmmm...
    how can it bee so many ways when there is only 10 ways to to put them on 3 hangers ... if position of shirt on hanger doesn't have any influence
    You wrote the number of ways to arrange five non-identical shirts on two identical hangers such that one hanger has three shirts and the other has two shirts. This is also just \binom{5}{2}.

    Edit: Alternatively, you wrote the ways to choose three shirts from a set of five non-identical shirts and hang them on three identical hangers, one per hanger. Unknown008's variation is that the hangers are considered non-identical.
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  7. #7
    MHF Contributor Unknown008's Avatar
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    Quote Originally Posted by undefined View Post
    I don't think it's clear what assumptions we're supposed to make. Supposing we can put multiple shirts on a hanger, a hanger can be empty, we must use all shirts, hangers are identical, and shirts are identical, there are 5 ways.

    {0,0,5}
    {0,1,4}
    {0,2,3}
    {1,1,3}
    {1,2,2}

    If the shirts are non-identical (which is a natural assumption), it becomes 41 ways

    {0,0,5} --- 1
    {0,1,4} --- 5
    {0,2,3} --- C(5,2)
    {1,1,3} --- C(5,3)
    {1,2,2} --- 5*C(4,2)/2

    and there are many variations of assumptions.
    Interesting... I didn't look at this question like this.

    Anyway, I tackled this question like the questions I did in the past. The question mentioned 'arrangements' and by experience, I've learned it the hard way that the assumptions are: the shirts are non identical and the order of the hangers is relevant. And lastly, only three shirts get a place on the hangers.

    Questions on combinations and permutations are not always explicit and can be interpreted in different ways...
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  8. #8
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by undefined View Post
    You wrote the number of ways to arrange five non-identical shirts on two identical hangers such that one hanger has three shirts and the other has two shirts. This is also just \binom{5}{2}.

    but that is also and \binom{5}{3} so it's 5 shirts , 3 hangers , and 10 different combinations (ways) to hang shirts (2 who don't have any hanger don't get to hang, or that's what i'm missing here )
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  9. #9
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by yeKciM View Post
    but that is also and \binom{5}{3} so it's 5 shirts , 3 hangers , and 10 different combinations (ways) to hang shirts (2 who don't have any hanger don't get to hang, or that's what i'm missing here )
    I added an "edit" to the post that you quoted, which addresses this. Maybe you're familiar with the notation that you wrote C(5,3) while Unknown008 wrote P(5,3).
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  10. #10
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by undefined View Post
    I added an "edit" to the post that you quoted, which addresses this. Maybe you're familiar with the notation that you wrote C(5,3) while Unknown008 wrote P(5,3).
    okay, but here (where i live) all problems that we get in that type of tasks is based on these that i wrote, and that's why I have assumed, and wrote that there Thanks for clarification
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