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Math Help - how many ways are there to add 3 numbers to get a bigger integer

  1. #1
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    Exclamation how many ways are there to add 3 numbers to get a bigger integer

    Hi there,
    I'm trying to achieve the number of ways you can add three numbers x, y and z to be equal to a given integer.

    For example, let's say the given integer is 10.
    so I need to know how many ways are there to add up three numbers x, y and z to be equal to 10?

    x+y+z = 10. The values of x, y and z would have to be either a positive whole number, or only a decimal number with .5

    Some of the ways are 1) x=0, y=0, z=10 2) x=0, y=5, z=5 3) x=4.5, y=4.5, z = 1. This list can go on, but will still be infinite given that u can only use either a positive whole number or a decimal number with only .5 such as 0.5, 5.5, 9.5 etc.

    Do you think is it possible? If so, is there a general way to achieve this for any given integer rather than just 10?
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  2. #2
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    Do you want the number of non negative integral solutions to, say,

    x_{1}+x_{2}+x_{3}=10?.

    If so, we can do a change of variables.

    Let y_{1}=x_{1}-1, \;\ y_{2}=x_{2}-1, \;\ y_{3}=x_{3}-1

    This gives us y_{1}+y_{2}+y_{3}=7

    \binom{7+3-1}{7}=\binom{9}{7}=36

    The number of integral solutions without restriction would be

    \binom{10+3-1}{10}=\binom{12}{10}=66
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  3. #3
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    Quote Originally Posted by dgmath View Post
    Hi there,
    I'm trying to achieve the number of ways you can add three numbers x, y and z to be equal to a given integer.
    For example, let's say the given integer is 10.
    so I need to know how many ways are there to add up three numbers x, y and z to be equal to 10?
    x+y+z = 10. The values of x, y and z would have to be either a positive whole number, or only a decimal number with .5
    Some of the ways are 1) x=0, y=0, z=10 2) x=0, y=5, z=5 3) x=4.5, y=4.5, z = 1. This list can go on, but will still be infinite given that u can only use either a positive whole number or a decimal number with only .5 such as 0.5, 5.5, 9.5 etc.
    There is one major fact you did not explain.
    Do you consider x=4.5,~y=4.5,~z=1 to be different from x=1,~y=4.5,~z=4.5.

    The answer to this depends upon knowing if these are the same or different.
    Once you tell us that then we can give a general solution for any N and any number of terms in the sum.
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  4. #4
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    Do you consider to be different from .

    Plato,
    yes, they're different from eachother.
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  5. #5
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    Quote Originally Posted by galactus View Post
    Do you want the number of non negative integral solutions to, say,

    x_{1}+x_{2}+x_{3}=10?.

    If so, we can do a change of variables.
    I'm not quite sure what you meant by that. Can you please rephrase it for me? Thanks.
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  6. #6
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    Quote Originally Posted by dgmath View Post
    yes, they're different from eachother.
    Then the answer is easy: \binom{20+3-1}{20}=\binom{22}{2}=\frac{22\cdot 21}{2}

    To explain. That is the number of non-negative integral solutions to x+y+z=20.
    If we divide each of those solutions by two we get one of your solutions.
    Example: 7+4+9=20 so 3.5+2+4.5=10

    So for a general solution.
    If N is a positive integer and we want to count the number of ways to add k non-negative integers or numbers with the decimal part .5 to get N, the idea is the same as above.
    \binom{2N+k-1}{2N}
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  7. #7
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    Quote Originally Posted by Plato View Post
    Then the answer is easy: \binom{20+3-1}{20}=\binom{22}{2}=\frac{22\cdot 21}{2}

    To explain. That is the number of non-negative integral solutions to x+y+z=20.
    If we divide each of those solutions by two we get one of your solutions.
    Example: 7+4+9=20 so 3.5+2+4.5=10

    So for a general solution.
    If N is a positive integer and we want to count the number of ways to add k non-negative integers or numbers with the decimal part .5 to get N, the idea is the same as above.
    \binom{2N+k-1}{2N}
    Yes! That is exactly what I was looking for. thanks bunch!
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