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I'm not professional mathematician but I do love a math problem - this one, however has me stumped.

I'm a UX Designer trying to figure out some guidelines for using tables in a page layout. The thing I want to know is how many possible combinations of cells across the width I can use to make up a 12 column table but only using column units equivalent to 1, 2, 3, 4, 6 and 12 columns - the whole number results of dividing the total as many ways as possible.

I know, for example that I can create a full width single column using a single 12-column cell. Or two equal columns using two 6-column cells... but after that it starts to get tricky.

I can make 3 columns using three 4-column cells but I can also make three columns using one 6-column cell and two 3-column cells.

And getting to four or more cells gets even more complex.

So, to sum up, I'd like to know if there is a way to work out how many possible combinations of the whole number divisions of 12 can be used to total 12 (regardless of addition order - so 6+3+3, 3+6+3 and 3+3+6 only count as one.)

Does that make any sense?

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Making Change for a Dollar (and other number partitioning problems) is a related question that provides a lot of background on how to solve this sort of problem. In your case, you want the coefficient of $x^{12}$ in the generating function

$$\frac1{(1-x)(1-x^2)(1-x^3)(1-x^4)(1-x^6)(1-x^{12})}\;,$$

for which Wolfram|Alpha yields $45$ (you need to press "more terms" twice).

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  • $\begingroup$ Thanks for the answer. I don't understand how that works but I'm going to keep figuring it out and try some other examples until I do! $\endgroup$ – Andrew Martin Jul 1 '15 at 11:56
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    $\begingroup$ @AndrewMartin: That question I linked to might be of help; otherwise feel free to come back and ask :-) $\endgroup$ – joriki Jul 1 '15 at 12:07

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