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I have equation: $x_1+...+x_{10}=30$
I should give the number of solutions if $x_i\ge 0 $ for $i\ge 3$ and:

1) $x_1\ge 3\ , \ 0 \le x_2\le 2 $

2) $0 \le x_1 \le 10 \ ,\ x_2 \ge 0$

3)$0 \le x_1 \le 10 \, \ 0 \le x_2 \le 10 $

In 1) I should consider for $x_1$ equel to 0,1 and 2, but still I need some formula. Can you help me with that?

user224301
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    Where are you stuck? Couple of hints: 1) $x_1\ge 3$ means $x_1-3\ge 0$, so you can change the variable $x_1' = x_1 - 3$. 2) $x_2\le 2$ is just the opposite inequality, so take all solutions and subtract those with $x_2\ge 3$ (which you should be able to count using the first hint). – zhoraster Sep 07 '15 at 13:05

1 Answers1

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Hints on 1) to put you on track.

  • under condition $x_1\geq3$ any solution of $x_1+x_2+x_3\cdots+x_{10}=30$ for nonnegative integers $x_1,x_2,\dots,x_{10}$ corresponds with a solution of $y_1+x_2+x_3+\cdots+x_{10}=27$ for nonnegative integers $y_1,x_2,\dots,x_{10}$. This in the understanding that $x_1=y_1+3$.
  • extra condition $x_2\leq2$ then tells us that it is enough to find the number of solutions of $y_1+x_3+\cdots+x_{10}=n$ where $n\in\{25,26,27\}$. These $3$ cases can be handled separately.
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drhab
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