hi 1975rachel

Welcome to the forum.

I agree with bobbym. I'll try to show you why this is impossible.

The first time it doesn't matter who sits with whom, so let's get them seated and then give everyone a number as follows.

1 2 3 4 5 6

7 8 9 10 11 12

13 14 15 16 17 18

18 20 21 22 23 24

25 26 27 28 29 30

Now who will person 1 sit with, the next time around?

No one from 2-6, so I'll choose someone from the next table, let's say person 7. As those six folk can shuffle their numbers around 7-12, I'm not losing any generality by saying 7.

Now who will 1 and 7 sit with?

It'll have to be someone who wasn't on either of the first tables so let's choose person 13. Aagin I'm not losing generality by picking that number (it could equally be 14 or 15 ...)

So who will 1, 7 and 13 sit with?

I'll have to choose from a table that hasn't been picked yet so I'll have person 18, ... and to speed things up I'll pick 25 next.

So I've got 1, 7, 13, 18, 25.

Now you might think I've fiddled it by choosing these numbers. But my point is, to avoid putting people with folk they've already met, you have to choose one person from each table.

So, for example, 1, 8, 15, 22, 30 is another way of picking five to go on the first table in round two.

But I have only picked five people and the table must seat six. So who else can I choose?

As I have already picked one person from each of the tables, there's no one left who I can pick without putting someone with someone they've already met.

Conclusion: It is impossible.

Bob