$$x_1+x_2+x_3+x_4+x_5+x_6<538|x_i>0$$ The inequality is strict. Apparently, there are $\binom{537}{6}$ solutions to this after adding a new variable, $x_7$, that serves as a balance. Could someone walk me through this step-by-step? I realize that, if the inequality is strict, then $x_7 > 0$. I'm confused about the logic of counting the number of integer solutions.
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2$\begingroup$ Look up stars and bars and search for that on this site. $\endgroup$ – Ross Millikan Sep 8 '20 at 14:12
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$\begingroup$ @RossMillikan Thanks for the guidance. I posted an answer $\endgroup$ – elbecker Sep 8 '20 at 14:26
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Imagine 538 placeholder spaces to put an object. First add a balance variable $x_7$ to make an equality. $$x_1+x_2+x_3+x_4+x_5+x_6+x_7=538|x_7>0$$ The restriction on this variable is that $x_7>0$ for the equality and original strict inequality to hold true. Now distribute 1 object each to variables ${x_1...x_7}$ due to the restriction that $x_i>0$. This leaves 531 objects to distribute among 7 variables. There are 6 dividers to partition the set into 7 parts, which means there are $531+6=537$ placeholders for either a divider or an object. The answer is therefore $537\choose{6}$.
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$\begingroup$ $x_7$ $\ge 0$, your answer is wrong $\endgroup$ – user800216 Sep 8 '20 at 16:33
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$\begingroup$ @leonard I don't see what is wrong with it since a strict inequality has to be maintained. could you explain in more detail please $\endgroup$ – elbecker Sep 8 '20 at 16:38
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$\begingroup$ i posted my answer .you can find what i meant. respects... $\endgroup$ – user800216 Sep 8 '20 at 16:46
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$\begingroup$ what you missed is that $x_7$ does not have to be strict $\endgroup$ – user800216 Sep 8 '20 at 16:50
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$\begingroup$ @elbecker: When you force $x_y$ to be positive and require that $x_1+\ldots+x_7=538$, you’re forcing $x_1+\ldots+x_6$ to be strictly less than $538$, thereby eliminating the solutions in which $x_1+\ldots+x_6=538$. You need to replace $538$ by $539$ if you want to require that $x_7$ be positive. Then your approach works and is at least arguably slightly simpler than elbecker’s. $\endgroup$ – Brian M. Scott Sep 8 '20 at 16:51
