Our little group here in the woods consists of four people. As I’ve been watching the various conversations develop, I’ve been observing how the social dynamics change, depending upon who is or isn’t present in any particular discussion.

Counting all the possibilities, I see that there are eleven possible conversational groupings among four people: (1,2) (1,3) (1,4) (2,3) (2,4) (3,4) (1,2,3) (1,2,4) (1,3,4) (2,3,4) (1,2,3,4).

Between any two people there is only one possible set of groupings, and between any three people there are four: (1,2) (1,3) (2,3) (1,2,3).

There turns out to be a simple formula based on the number of people, which is how I know that for five people there are 26 possible conversational groups, for six people there are 57, and for seven people there are 120. **

I will leave it up to you to figure out what the formula is.

*** My original post contained typos for some of these numbers, which Sharon and Logan kindly corrected.*

Why am I getting 26 for 5 people and 57 for 6?

http://www.wolframalpha.com/input/?i=2%5En-n-1+for+n%3D1+to+10

2^n (subsets) -n (no talking to yourself) -1 (if a conversation is held by no one does it really take place?)

I came about it from a less elegant approach than Adam, but it gives the same values:

n n

SUM (n choose k) = SUM (n!/k!(n-k)!

k=2 k=2

(that is, summing over all sizes of groups of at least 2, the number of ways to form a group of that size given n people).

However, both my formula and Adam’s give 26 for n = 5, and 57 for n = 6. Ken?

Yuck, that didn’t format well. The second formula is

n

SUM (n!/k!(n-k!))

k=2

@Logan: That’s funny, I must have made the same arithmetic errors when I checked my results. I definitely remember coming up with 47 (64-6 = 48 instead of 58).

Yes, Adam’s formulation — 2^n – n – 1 — is the one I was thinking of.

Thanks Sharon and Logan. I typed the original post too fast, and didn’t notice I had entered the wrong values. I’ve corrected the post so as not to completely confuse future readers.