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  1. #1
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    Integer Soultion to an equation...

    How many integer solutions (x1; x2; x3) are there to the equation :-

    x1 + x2 + x3 = 25


    with all xi satisfying 0 < xi < 10?
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    Quote Originally Posted by aamiri View Post
    How many integer solutions (x1; x2; x3) are there to the equation :-

    x1 + x2 + x3 = 25


    with all xi satisfying 0 < xi < 10?
    Maybe not the most elegant, but:

    The lowest possible for x1 is 5, forcing (x2,x3)=(10,10).

    Fix x1=6 then it's either (x2,x3)=(9,10) or (10,9)

    Fix x2=7 then it's either (x2,x3)=(8,10) or (10,8) or (9,9)

    As you can see there really aren't that many options if you continue in this manner.
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    ^^ Thanks for taking out time to reply.

    So, in this case , do I just keep going in the same manner as you did?

    Is there any other way of doing this ?
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    Quote Originally Posted by aamiri View Post
    ^^ Thanks for taking out time to reply.

    So, in this case , do I just keep going in the same manner as you did?

    Is there any other way of doing this ?
    Well for these numbers I think it's fine to continue in the same manner, it's quick enough. For an extended problem like

    x1 + x2 + ... + x7 = 39

    one approach if you are allowed a computer is to solve recursively. For other ways to solve, maybe someone else will chime in.
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    Quote Originally Posted by aamiri View Post
    How many integer solutions (x1; x2; x3) are there to the equation :-

    x1 + x2 + x3 = 25


    with all xi satisfying 0 < xi < 10?
    A generating function solution requires the least thought, in my opinion.

    Suppose, more generally, we want to count the integer solutions to
    x_1 + x_2 + x_3 = r
    where 0 \leq x_i \leq 10;
    let's say the number of solutions is a_r.
    Then a_r is the coefficient of x^r in
    f(x) = (1 + x + x^2 + \dots + x^{10})^3.

    From here on it's just a matter of manipulating series:
    f(x) = \left( \frac{1-x^{11}}{1-x} \right) ^3
    = (1- x^{11})^3 (1-x)^{-3}
    = (1 - 3 x^{11} + 3 x^{22} - x^{33}) \sum_{i=0}^{\infty} \binom{3+i-1}{i} x^i

    Picking out the coefficient of x^{25} from this expression, we can see it is

    a_{25} = \binom{27}{25} - 3 \binom{16}{14} + 3 \binom{5}{3}.

    I think a solution via inclusion/exclusion is also possible (I haven't worked it out), but I think it will lead to the same expression for the answer. Only you will have to think harder. :-P
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