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"Find the number of solutions of $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 34$ in positive even integers not exceeding 10."

Answer: $\binom{16}{5} - 6\binom{11}{5} + 15\binom{6}{5}$

According to the book "Maths of Choice, the answer was derive from the following: "the question amounts to asking for the number of solutions of $y_1 + y_2 + ... + y_6 = 17$ in positive integers not exceeding 5, because any even integer $x_1$ can be written as $2y_1$, where $y_1$ is again an integer.

Below is how I think the solution is derived; however, I'm not sure where the 5s from $\binom{17-5-1}{6-1}$ and $\binom{17-5-5-1}{6-1}$ are from. Are they the positive even integers between 1 to 10 or "17 in positive integers not exceeding 5"?

n: $\binom{17-1}{6-1} = \binom{16}{5}$

n(a) = $\binom{17-5-1}{6-1} = \binom{11}{5}$

n(ab) = $\binom{17-5-5-1}{6-1} = \binom{6}{5}$

Kindly advise

ManOnTheMoon
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  • See this answer for a step by step approach of the generic problem. – user2661923 May 23 '22 at 08:39
  • @user2661923, thank you for the reply. I am actually asking more of whether the author is asking for "positive even integers" or "positive integers". – ManOnTheMoon May 23 '22 at 08:49
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    The author first transformed the problem of finding the number of solutions to the equation $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 34$ in positive even integers not exceeding $10$ to finding the number of solutions to the equation $y_1 + y_2 + y_3 + y_4 + y_5 + y_6 = 17$ in positive integers not exceeding $5$. Therefore, the author is subtracting cases in which one or more of the variables in the second equation I wrote exceeds $5$. – N. F. Taussig May 23 '22 at 08:54

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