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Math Help - combinatorics (numbers whose digits add up)

  1. #1
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    combinatorics (numbers whose digits add up)

    Ive been stuck on this problem for a while.
    We're supposed to find the number of numbers whose digits add up to 20 between 0 and 999,999.
    I've done it with adding up to 15, which is just 20C5 - 6 * 10C5 but now I am stuck.
    I know that the total numbers that add up to 15 including violators is 25C5 but I can't figure out how to handle the incorrect cases.
    I thought about just checking for one of the numbers > 10 and then just subtract the case if two numbers were 10 exactly, but that didnt' work.
    My incorrect equation for that is 25C5 - 6 * 15C5 - 6C2.

    Any thoughts?

    Thanks
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  2. #2
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    Re: combinatorics (numbers whose digits add up)

    Quote Originally Posted by adameast View Post
    Ive been stuck on this problem for a while.
    We're supposed to find the number of numbers whose digits add up to 20 between 0 and 999,999.
    I've done it with adding up to 15, which is just 20C5 - 6 * 10C5 but now I am stuck.
    I know that the total numbers that add up to 15 including violators is 25C5 but I can't figure out how to handle the incorrect cases.
    I thought about just checking for one of the numbers > 10 and then just subtract the case if two numbers were 10 exactly, but that didnt' work.
    My incorrect equation for that is 25C5 - 6 * 15C5 - 6C2.
    I worked this problem using a generating function.
    As you can see 35127x^{20} appears, giving the answer.

    Notice that your work \binom{25}{5}-6\cdot\binom{15}{5}{\color{red}+}\binom{6}{2}=3512  7 gives the same answer.
    NOTE the {\color{red}+}.

    Please tell us what is the reasoning behind your approach.
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