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**adlady****Member**- Registered: 2012-11-07
- Posts: 2

I can't figure out how to do this. We don't want to sit at a table with someone we have already visited with. We know we won't get to see all the people. Just need 3-4 rounds. Can someone help?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,890

Hi;

Give each person a different number from 1 to 42.

First day:

Second day:

Third day:

Fourth day:

Each row represents a table. For instance on the first day table 3 has persons 13,14,15,16,17 and 18 eating together. Fourth day, table 7 has 11,15,19,29,33 and 37 eating together.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**adlady****Member**- Registered: 2012-11-07
- Posts: 2

Thank you very much. I don't know how you did it. It is appreciated.

*Last edited by adlady (2012-11-08 03:56:10)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,890

Hi adlady;

Welcome to the forum and save me some chicken!

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**1975rachel****Member**- Registered: 2012-11-21
- Posts: 3

Hello, I really hope you can help me.

I have the same problem but it is for 5 tables of 6 people and I need just to rotate it for 3 courses. The idea being that no one sits with the same person twice.

Please please could you help me as I am losing the will to live.

Thanking you in advance.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,890

Hi 1975rachel;

I am sorry I do not know of any structure of 5 tables of 6 people for any number of courses. Not all structures are possible. The best I can do is 6 tables of 5 people each for 3 courses.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**1975rachel****Member**- Registered: 2012-11-21
- Posts: 3

Hi bobbym

Many thanks for coming back to me.

Unfortunately we are stuck with this seating plan!

Nevermind, thanks for your help.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,890

Hi;

Sorry that I can not help but some of them I just do not know the answer to or even if there is an answer.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,006

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

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**1975rachel****Member**- Registered: 2012-11-21
- Posts: 3

Hi Bob,

You are right and at least it means I wasn't being that stupid!

Thank you ever so much for your time and help, it is very kind of you.

I have devised a scheme which sort of works, there is some cross over, but it will do.

Thank you again

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,890

Hi;

Would you consider allowing me to see your solution?

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

**Online**