# Number of unordered combinations of N-sized subsets of an m*N-sized set

I have an array of 24 batteries in my off-grid solar power system. They are arranged in a series of 4 blocks having 6 parallel-connected batteries in each. So, in my case, N=6 and m=4.

The batteries slightly vary by state of health, which is reflected in their internal resistance. This causes each of the 4 serial blocks to have slightly different resistance, which in turn causes the blocks to get charged to unequal voltage which is not ideal (some of the 4 blocks get charged to a higher voltage than the other, although they all get the same current). Given that I can measure internal resistance of each battery individually, I am about to come up with the best possible combination where the 4 blocks have the closest resistance possible.

Now, calculating resistance for a given combination is easy and is not a question.

The question is how to get all those (unordered) combinations? How many are there, and how to iterate over all of them? (so that I could write a script that would just try them all and return the best combination)

Say if N was 2 and m was 2, the 4 batteries were labeled from A to D, there would have been only 3 unordered combinations: AB+CD, AC+BD and AD+BC.

For N=3 and m=2 we get 10 (unless I have missed some):

1. ABC+DEF
2. ABD+CEF
3. ABE+CDF
4. ABF+CDE
5. ACD+BEF
6. ACE+BDF
7. ACF+BDE
8. ADE+BCF
9. ADF+BCE
10. AEF+BCD

So, where m=1 there is only one unordered combination no matter how big N is. But what is the equation?

## 1 Answer

There are $$24 \choose 6$$ ways to choose the first pack, $$18 \choose 6$$ ways to choose the second, $$12 \choose 6$$ to choose the third, and $$6 \choose 6$$ ways to choose the last. We have overcounted by a factor $$4!$$ because we could choose the same four packs in any order. This gives $${24 \choose 6}{18 \choose 6}{12 \choose 6}{6 \choose 6}\frac 1{4!}=\frac {24!}{6!^4\cdot 4!}\approx 9.62\cdot 10^{10}$$ Generally is is $$\frac{(mN)!}{(N!)^m\cdot m!}$$

• Thank you! How do you call that construct with two numbers one above other in brackets? – Greendrake Apr 21 at 14:09
• $N \choose m$ is the number of combinations of $m$ things out of $N$ without regard to order. It equals $\frac {N!}{m!(N-m)!}$ – Ross Millikan Apr 21 at 14:11