Given a set of N resistors, not guaranteed identical, how many different arrangements using k resistors can be made?

75 Views Asked by At

I'm trying to develop a program that evaluates all possible arrangements (combinations of series and parallel) of k resistors from a set.

From what I've read, if the resistors were identical values, then this would just be a permutation formula. But what if the k resistors have different values? Say I don't care if any arrangements end up having the same net resistance as each other, I just want to know the total number of ways I can arrange R1, R2, ... Rk.

Ignoring the set for the moment (or setting N = k), I get that when k =1, there's only one arrangement (obviously). When k=2, there are 2 (both in series, both in parallel).

When k = 3, I get 8:

1 x all three in series

1 x all three in parallel

3 x 1 in parallel with 2 in series, and

3 x 1 in series with 2 in parallel

When k becomes 4 or above, this becomes a lot more complicated. 3 series resistors can be parallel with 1, or 1 can be in series with 3 parallel, or 2 parallel resistors can be in series with 2 other parallel resistors, or 2 series can be in series with 2 parallel, etc.

Does anybody have a solution for this problem, or know of any statistical formula that would help me?

EDIT: after a quick and dirty attempt to sketch out all possible arrangements for 4 resistors, I counted 52 different arrangements. As best as I can explain:

1 x all 4 in series

1 x all 4 in parallel

4 x 1 in series with 3 in parallel

12 x 1 in series with 1 resistor in parallel with 2 others in series

6 x 2 in series, in series with 2 in parallel

3 x 2 in parallel in series with 2 in parallel

3 x 2 in series parallel with 2 in series

6 x 2 in series in parallel with 2 in parallel

4 x 1 in parallel with 3 in series

12 x 1 in parallel with (1 in series with 2 in parallel)

That's probably not very easy to follow, but I'm pretty sure I'm correct. So now I have for N=k,

k = 1: y = 1

k = 2: y = 2

k = 3: y = 8

k = 4: y = 52

k = 5: y = ??