# Calculating the total number of sets from multiple choices

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.

• $n!\over {k!(n-k)!}$ – user451844 Aug 4 '17 at 13:58
• 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 – Ethan Bolker Aug 4 '17 at 13:58

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, $2!=2$.