I'm trying to calculate the number of potential options in creating characters for Dungeons and Dragons. Most of this is simple multiplication, but sometimes a character is allowed to choose two of something with no duplication and order doesn't matter.

Half-Elves can choose to increase two attributes from Strength, Dexterity, Constitution, Intelligence, and Wisdom. I thought the answer was number of initial choices times number of remaining choices (5x4=20), but that implies that it matters which one is selected first, which is irrelevant. I wrote it out, and discovered there are only 10 outcomes.

So then I thought the answer as number of initial choices times number of selections (5x2=10).

But then I increased the number of choices to six and wrote it out, and the answer is 15.

This is going to come up again with skills and weapons. So how do I calculate the number of potential combinations?

Thank you.

  • $\begingroup$ $n!\over {k!(n-k)!}$ $\endgroup$
    – user451844
    Aug 4, 2017 at 13:58
  • 4
    $\begingroup$ You're counting combinations for each attribute, and you are on the right track. The formulas you need are here: en.wikipedia.org/wiki/Combination $\endgroup$ Aug 4, 2017 at 13:58

1 Answer 1


You want the binomial coefficient, $\binom{5}{2}$ (read as "5 choose 2", which you can also type into google). As the name suggests, it is the number of ways you can choose 2 items from 5 items, no repeats, order doesn't matter.

The forumula for the binomial coefficient is $\displaystyle\binom{n}{k}=\frac{n!}{k!(n-k)!}$.

What you were doing, $5*4$, is essentially $\dfrac{n!}{(n-k)!}$ with $n=5$ and $k=2$. To get the binomial coefficient, you divide again by $k!$. This basically accounts for repeats where the order is different but the items are the same. In this case, you are choosing two items, so you could have $\{\!\{a,b\}\!\}$ and $\{\!\{b,a\}\!\}$, which are the same thing. The number of ways you can order $k$ items, or the number of permutations, is $k!$. In this case, $2!=2$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .