# How to sit to meet everyone

1. May 15, 2005

### Xamfy19

Hello All;
12 people use a dining room with 3 tables. Each table can sit 4 person. How to arrnage the seatings in 5 days so that each person can share the table at dinner with each other person?

thanks alot,,, :uhh:

2. May 16, 2005

### honestrosewater

A general approach doesn't spring to mind but notice that if you arrange the people into pairs, each pair has five other pairs to sit with (i.e. one pair each day). Say you pair them (1, 2), (3, 4), (5, 6), ... (11, 12).
$$\begin{array}{|c|c|c|}\hline \mbox{Day}&(1, 2)\ \mbox{sits with}&(3, 4)\ \mbox{sits with} \\ \hline 1&(3, 4)&(1, 2) \\ \hline 2&(5, 6)&(9, 10) \\ \hline 3&(7, 8)&(11, 12) \\ \hline 4&(9, 10)&(7, 8) \\ \hline 5&(11, 12)&(5, 6) \\ \hline \end{array}$$
You can just complete the table by hand. You may be after a better way to do this, but it does work ;)

3. May 16, 2005

### matt grime

That's quite a good way. ANother way to explain it is, after rewriting it as a problem with 6 people to meet on five days (thinking of a pair as a person) sit them down as pairs

1 2

3 4

5 6

then keeping 1 fixed rotate the other five pairs anticlockwise (or clockwise) once each of the 5 days.

This looks very much like a problem adapted from Bridge seeing as it can be done with pairs.

4. May 16, 2005

### honestrosewater

So I can't help but wonder how many solutions there are (for pairings). 6C2 = 15 unordered pairs. But how do you split those up into 5 sets of 3, all disjoint?

At the start of each round of choosing 3 pairs, each individual (i.e. 1, 2, 3, 4, 5, 6) is in n of the p pairs. For the first round n = 5 and p = 15, and each round you subtract 1 from n and 3 from p. Figuring out how each round of choosing works is trickier; I see what's at work, but it's not clear how it's working. Every time you choose a pair, all pairs containing those individuals cannot be chosen that round, thus for every other individual, 2 pairs containing it are excluded. However, the choices are further restricted in a way I can't really explain. For instance, say {1, 4} is the first choice in round 3, leaving {2, 3}, {2, 6}, {3, 5}, {3, 6} -but I can only choose {2, 6} or {3, 5}.
Anyway, working it through, I get:
Round 1: 15C1 * 6C1 * 1C1 = 90
2: 12 * 4 * 1 = 48
3: 9 * 2 * 1 = 18
4: 6 * 2 * 1 = 12
5: 1
And I think I just sum the rounds: 169 solutions.? That seems like too many.

Last edited: May 16, 2005
5. May 16, 2005

### Xamfy19

thanks alot, gentlemen!