How many integer solutions are there to the inequality $$y_1+y_2+y_3+y_4\lt184$$with $y_1\gt0$, $0\lt y_2\le10$, $0\le y_3\le17$, and $0\le y_4\le 19$?
How do we solve this using Incl. Excl. Principle?
I can see it clearly using Generating Function
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infinity
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Did you mean to mix strict and non-strict orders? – copper.hat Oct 11 '13 at 05:42
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I gave you a very large hint in my answer to your previous question. I don’t think that I can say much more than that without simply doing the problem for you, which I’m not willing to do — especially since I did already completely solve a similar problem for you. – Brian M. Scott Oct 11 '13 at 05:44
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@copper.hat: Yes: it’s a scanned image of the source of the question. I imagine that whoever posed the question wanted to force the student to deal with a few minor complications. – Brian M. Scott Oct 11 '13 at 05:46
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@BrianM.Scott: Thanks! – copper.hat Oct 11 '13 at 05:48
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@BrianM.Scott I solved it. I just wanted to check the answer. I am sorry. – infinity Oct 11 '13 at 06:14
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"I solved it. I just wanted to check the answer." Then show what you did--otherwise how to check that it is correct? – Did Oct 11 '13 at 07:47
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@infinity it's good that you solved it and want to check, just post that and more people will be willing to say if you made a mistake or not – DanielV Oct 11 '13 at 07:56
1 Answers
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Here is one tedious way: $S=\sum_{i_4=0}^{18} \sum_{i_3=0}^{17} \sum_{i_2=1}^{10} (183-(i_2+i_3+i_4))$.
This works because $18+17+10 < 183$.
Since I need to procrastinate on some urgent but even more tedious work...
\begin{eqnarray} S &=& 19 \cdot 18 \cdot 10 \cdot 183 -19\cdot 18 \sum_{i_2=1}^{10} i_2 -19 \cdot 10 \sum_{i_3=0}^{17} i_3 - 18 \cdot 10 \sum_{i_4=0}^{18} i_4 \\ &=& 625860 - 342 \sum_{i_2=1}^{10} i_2 -190 \sum_{i_3=1}^{17} i_3 - 180 \sum_{i_4=1}^{18} i_4 \\ &=& 625860 - 342 \frac{1}{2}10 \cdot 11 -190 \frac{1}{2}17 \cdot 18 - 180 \frac{1}{2}18 \cdot 19\\ \\ &=& 547200 \end{eqnarray} (Assuming I calculated correctly.)
copper.hat
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