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How many nonnegative integer solutions are there to the pair of equations \begin{align}x_1+x_2+\dots +x_6&=20 \\ x_1+x_2+x_3&=7\end{align}

How do you find non-negative integer solutions?

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    $\begingroup$ The second factor in your answer should be based on the number of solutions of $x_4+x_5+x_6=13$. $\endgroup$ – paw88789 Oct 4 '14 at 23:38
  • $\begingroup$ my bad. Is it correct now? $\endgroup$ – Lil Oct 4 '14 at 23:42
  • $\begingroup$ I think it's ok now. $\endgroup$ – paw88789 Oct 4 '14 at 23:43
  • $\begingroup$ so is it always arranged where the first number is what the equation equals and then you add one less then the number of terms? $\endgroup$ – Lil Oct 4 '14 at 23:44
  • $\begingroup$ Saying 'always' is a dangerous thing in math. Subtle differences in wording can change answers quite a bit. But in your case you can use a 'stars and bars' argument. See for instance en.wikipedia.org/wiki/Stars_and_bars_(combinatorics) $\endgroup$ – paw88789 Oct 4 '14 at 23:47
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You are correct. You can also think of it in terms of permutations.

The number of non-negative integer solutions of $x_1+x_2+x_3=7$ is the number of permutations of a multiset with seven $1$'s, and two $+$'s. This is $$\frac{9!}{7!\ 2!}.$$ Similarly, the number of non-negative integer solutions of $x_4+x_5+x_6=13$ is the number of permutations of thirteen $1$'s, and two $+$'s. This is $$\frac{15!}{13!\ 2!}.$$

This is why the first number in your combination is what the variables equal, and the second is "one less" the amount of variables, since you're permuting the $+$'s.

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  • $\begingroup$ similarly if I had x4+x5=9 it would be (10!/1!9!)? $\endgroup$ – Lil Oct 5 '14 at 0:00
  • $\begingroup$ Yup. Nine $1$'s and one $+$ are $10$ things in total. You have to $\textbf{choose}$ one spot out of ten for the $+$, which is just $C(10,1)$. Of course this also $C(10,9)$. So yes you are correct! $\endgroup$ – Andrey Kaipov Oct 5 '14 at 0:09

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