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Combinations problem

  1. May 29, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the number of solutions to the equation x + y + z + t = 16, where x, y, z, and t are:

    i) odd integers
    ii) even integers

    2. Relevant equations

    N/A

    3. The attempt at a solution

    I know the total number of solutions is 19 choose 3. So I'm thinking I can obtain the answer for i) by subtracting the universal set by the set where at least one of the variables is an even integer.

    I cannot figure out how to calculate the number of solutions where at least one variable is even.
     
  2. jcsd
  3. May 30, 2007 #2

    VietDao29

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    Just one question, can x, y, z, and t be the same? If they cannot be the same, then for odd x, y, z, t, there can only be one set of solution (1, 3, 5, 7). Then, you can start switching them a little bit, and come to the answer.

    ----------------------

    If they can be the same (and I suppose this is the case the problem wants us to do), then, here's my approach, require a little bit working. But, well, it works.

    x + y + z + t = 16

    Now, let M = x + y, and N = z + t, we have:
    M + N = 16

    This can be done easier, since it's the sum of two numbers, instead of the original four numbers.

    Since the sum of two odd integers is even, we have: M, and N must be even. The minimum value for, x, y, z, and t is 1, so the minimum value for M, and N is 2. The maximum value for M, and N is 9 + 9 = 18, i.e, we have:
    [tex]2 \leq M, N \leq 18[/tex]

    You'll have 7 ways to choose M, and N
    (2, 14), (4, 12), (6, 10), (8, 8), (10, 6), (12, 4), (14, 2)

    If M = 2, there'll be 1 way to choose x, and y: x = y = 1.
    If M = 4, there'll be 2 ways: (1, 3), and (3, 1)
    If M = 6, there'll be 3 ways.
    M = 8, 4 ways.
    M = 10, 5 ways
    and M = 12, 4 ways.
    M = 14, 3 ways.

    Do the same for N.

    Now, if M = 2, then N = 14, so there'll be 1 * 3 = 3 ways to choose x, y, z, t.
    M = 4, N = 12, there'll be 2 * 4 = 8 ways.
    M = 6, N = 10, there'll be 3 * 5 = 15 ways.
    M = 8, N = 8, 4 * 4 = 16 ways.
    M = 10, N = 6, there'll be 3 * 5 = 15 ways.
    M = 12, N = 4, there'll be 2 * 4 = 8 ways.
    M = 14, N = 2, there'll be 1 * 3 = 3 ways.

    Sum all the above, and you'll have the answer.
    3 + 8 + 15 + 16 + 15 + 8 + 3 = 68 ways.

    Can you get it? :)

    You can do exactly the same to the case, when x, y, z, and t are even.


    It's a little bit large, I think. Are you sure that's the answer?
     
    Last edited: May 30, 2007
  4. May 30, 2007 #3
    I probably should have included this link: http://mathforum.org/library/drmath/view/68607.html

    Your answer does seem to be a bit small though, only 68 solutions with all odd integers but it seems right, thanks.

    EDIT: Does your answer take into account negative numbers as well? And I believe zero is an integer as well...

    EDIT2: Actually now that I think about it, wouldn't there be infinite answers since negatives are included?
     
    Last edited: May 30, 2007
  5. May 30, 2007 #4

    VietDao29

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    If I take into account negative number, then it would be infinity combinations possible.

    Say x = -1 000 000 000.
    y = 999 999 999.
    z = 1.
    t = 16.

    Then, subtract 2 from x, and add 2 to z, we have a new solution.

    x = -1 000 000 002.
    y = 999 999 999.
    z = 3.
    t = 16.

    EDIT:
    Yes, of course. :)

    EDIT 2:
    0 is an even integer, not odd integer.

    Well, do you understand what the site says? The problem you problem provided, and the problem the site was trying to solve is not the same. =.="
     
    Last edited: May 30, 2007
  6. May 30, 2007 #5
    Ah thanks, Also I left this one out initially:

    x + y + z + t = 14, where x, y, z, and t are integers that is:
    i) x >= 3, y>= -1, z>= -4, t>=1

    Would this just be 18 choose 3?
    3 + -1 + -4 + 1 = -1, so add 1 to 14 = 15
    15 + 3 (The dividers) = 18 and then choose 3...
     
    Last edited: May 30, 2007
  7. May 30, 2007 #6

    VietDao29

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    Is this homework? It looks complicated to me, at least.
    What have you tried?
     
  8. May 30, 2007 #7
    Yes it is homework, due in about 15 minutes :P The only thing I can think of and have tried is the above, resulting in 18 choose 3. I'm not sure if my thinking is correct in saying that if the sum of the initial groups is -1, then I can add one to the total... I'm mainly trying to apply what I read from the link I posted a few posts back.
     
  9. May 30, 2007 #8

    siddharth

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    Gold Member

    Imagine that you 14 "1's" and you have to separate them into 4 "bins"

    For example, one way to do this is

    Bin 1 - Eight 1's
    Bin 2 - Three 1's
    Bin 3 - Two 1's
    Bin 4 - One 1

    So, if you indicate the separation by a " | ", it's
    11111111 | 111 | 11 | 1

    Now, can you see that the total number of ways in which you can arrange fourteen 1's into 4 bins is the same as the number of distinct arrangements of the sequence of 14 1's and 3 | ? (Which is [tex]^{17}C_14[/itex])

    But, this is exactly the same as the number of solutions to the equation
    x+y+z+t=14, where x,y,z,t are integers greater than or equal to 0.

    So, can you take it from here? How would you try simplyfying your conditions on x,y,z,t? (Hint: Introduce new variables)
     
    Last edited: May 30, 2007
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