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?


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


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

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.