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Math Help - w + x + y + z = 9

  1. #1
    Super Member Deadstar's Avatar
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    w + x + y + z = 9

    How many solutions in non-negative integers are there to the equation w + x + y + z = 9?

    How many solutions in positive integers?

    Think i figured out the first part which i got to be {12 \choose 3} = 220

    But how do i calculate the second part? So far im at 220 - (number of solutions with three 0's in it) - (number of solutions with two 0's in it) - (number of solutions with one 0 in it) = 220 - (4) - (6 x 8) - ?

    Is there a formula or is it just a counting exercise for the second part?
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  2. #2
    Eater of Worlds
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    Non-negative and positive. Isn't that pretty much the same thing?. Or do you just mean using 1 thrugh 9 instead of 0 through 9?.

    Then you would have: \left(\sum_{k=1}^{9}x^{k}\right)^{4}

    The coefficient of x^9 is 56, so there are 56 arrangements of the digits 1-9 that sum to 9.

    You could use the generating function, \left(\sum_{k=0}^{9}x^{k}\right)^{4}

    Then look at the coefficient of the x^9 term. Which is indeed 220.

    Suppose you wanted to do w+x+y+z=9, but only use the numbers 1 through 5

    Then \left(\sum_{k=1}^{5}x^{k}\right)^{4}

    The coefficient of x^9 is 52, so there are 52 ways to sum to 9 using the digits 1 through 5.
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  3. #3
    Junior Member pinion's Avatar
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    Quote Originally Posted by Deadstar View Post
    How many solutions in non-negative integers are there to the equation w + x + y + z = 9?

    How many solutions in positive integers?
    which question includes 0 in the set of integers?

    does your count include each of these?
    9+0+0+0=9
    0+9+0+0=9
    0+0+9+0=9
    0+0+0+9=9

    from the wording of the question, i would guess that the first question includes 0, and that your counts do allow the above. also, i would assume that each {w,x,y,z} can only be in {solutions} once. if these assumptions are correct, consider this:

    for each of these (you will see that this list has a spill-over pattern)
    9+0+0+0=9
    8+1+0+0=9
    7+2+0+0=9
    7+1+1+0=9
    6+3+0+0=9
    6+2+1+0=9
    6+1+1+1=9
    5+4+0+0=9
    5+3+1+0=9
    5+2+2+0=9
    5+2+1+1=9
    4+4+1+0=9
    4+3+2+0=9
    4+3+1+1=9
    3+3+3+0=9
    3+3+2+1=9

    you must find all the w, x, y, z permutations of the 4 numbers.
    i.e.
    6+3+0+0=9
    6+0+3+0=9
    6+0+0+3=9

    0+6+3+0=9
    0+6+0+3=9
    3+6+0+0=9

    0+0+6+3=9
    3+0+6+0=9
    0+3+6+0=9

    3+0+0+6=9
    0+3+0+6=9
    0+0+3+6=9

    as to finding some simple logic/equation to arrive at your answer, i don't know it off the top of my head, but i imagine that if there is an elegant solution, you will find it by studying the patterns in the permutations.

    i believe this question should have you look at and understand said patterns rather than have you function as a calculator, so if you're looking for a quick answer, i probably have not helped you here, but if you're looking to understand what it is you're doing, have at it!
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  4. #4
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    Quote Originally Posted by Deadstar View Post
    How many solutions in non-negative integers are there to the equation w + x + y + z = 9?

    How many solutions in positive integers?
    We can think of this as,
    (w-1)+(x-1)+(y-1)+(z-1) = 9.
    Where w,x,y,z are positive integers.
    This is equivalent to,
    w+x+y+z = 13.

    There are, {{13}\choose 3} ways to solve this.
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  5. #5
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    There is a conceptual mistake in the previous post.
    It should be \left( {w + 1} \right) + \left( {x + 1} \right) + \left( {y + 1} \right) + \left( {z + 1} \right) = 9 giving w+x+y+z=5 which has { 8 \choose 3} many positive integer solutions. I have often told classes to think of it as putting a one into each variable and then solving as usual.

    TPH’s answer gives the number of integer solutions in which the solutions are \ge -1.
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  6. #6
    Super Member Deadstar's Avatar
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    not quite sure i understand your post plato... {8 \choose 3} only gives 56 possible values. Is that for the second part? not including any 0's. I thought the formula {{n + k -1} \choose {n - 1}} solved these for (in this case) w,x,y,z \ge 0 which i thought would be {{4 + 9 - 1} \choose {4 - 1}} = {12 \choose 3}...
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  7. #7
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    Quote Originally Posted by Deadstar View Post
    not quite sure i understand your post plato... {8 \choose 3} only gives 56 possible values. Is that for the second part?
    That is the answer to this question.
    "How many solutions in positive integers?"
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