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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,786

I know. I think that she was a moron but she was right about me too.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**TracyBidwell****Member**- Registered: 2013-12-18
- Posts: 3

ShivamS wrote:

It's likely due to the fact that when you google "is 0 a rational number?", the first creditable result is from MIF.

Yup! Now, nothing is possible you can find the answer immediately in a blink of an eye. So if you have any fact to find the first thing to do is search in google.

Habeeb Akande

If you are not working towards something, your life will end with nothing.

― Habeeb Akande

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**allan1085****Member**- Registered: 2013-12-19
- Posts: 14

There are 4 progressive dinners, 5 groups of 4 couples each (20 couples alltogether). Each dinner has 4 courses. Using each couple as a number from 1 to 20, how can it work out so that each couple (number) does only 1 course and is never with any other couple more than once?

*Last edited by allan1085 (2013-12-19 10:11:38)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,786

Hi;

As far as I know 5 groups of 4 couples as you describe is not possible. This is the closest answer.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**allan1085****Member**- Registered: 2013-12-19
- Posts: 14

Hi bobbym, I guess I didn't explain myself very well. There are only 4 dinners. 5 groups of 4 each. The numbers must be scrambled so that each course is by a different couple. For example, you have #1 serving the 1st course each dinner.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,786

I probably can permute those first numbers, but if days = dinners then it is impossible to have 4.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**allan1085****Member**- Registered: 2013-12-19
- Posts: 14

bobbyum, Last year I was able to work it out for 16 couples, 4 to a group, 4 dinners. This year we added 4 more couples, but still only 4 dinners having 5 groups for each dinner. Yes, days=dinners.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,786

Not everyone of these progressive dinners has a solution, most do not. I am looking at the charts for all known solutions. 16 couples I believe is possible. I am seeing no solution for 20 couples, 4 dinners, 5 groups.

Now all of these progressive dinners have the rule that each couple eats with every other couple once and only once. If you want to relax the criterion and have the couples not eat with every other one then more solutions are possible.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**allan1085****Member**- Registered: 2013-12-19
- Posts: 14

bobbym, Yes, your suggestion is fine. Any solutions you can give me would be gresatly appreciated.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,786

Hi;

Since you require 4 dinners that means each couple can see 3 new couples per dinner. 4 x 3 = 12 couples maximum for the 4 dinners. That means each couple ideally will miss eating with 7 couples.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**allan1085****Member**- Registered: 2013-12-19
- Posts: 14

bobbym, No, there are still 20 couples. However, if some of the couples see each other more than once, than so be it. They just can't serve the same course more than once. I really appreciate ypour help.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,786

Look at couple #1. Take the first 4 dinners 1, 2, 3, 4.

1 eats with 2,3,4,5,6,9,10,13,15,16,18,20

Notice that 7,8,11,12,14,19 and 17 are missing.

1 ( and everybody else too ) will always be missing at least 7 people. This is because each dinner he eats with 3 different people. 4 x 3 = 12. 19 - 12 = 7.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**allan1085****Member**- Registered: 2013-12-19
- Posts: 14

bobym, The problem is that you have #1 serving the 1st course of 5 dinners. I am trying to get it so that no couple serves the same course more than once. If some couples are together more than once, than so be it. They just can't serve the same course more than once.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,786

I am working on that, I will post an answer as soon as I get it.

The solution may take awhile.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**allan1085****Member**- Registered: 2013-12-19
- Posts: 14

thanki you

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,786

How does this look?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**allan1085****Member**- Registered: 2013-12-19
- Posts: 14

thank you bobbym, you have been a great help.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,786

Hi;

Let me know if you need any adjustments. I did it by hand so there may still be inconsistencies in the arrangements.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,786

Hi;

Please do not use the report button to contact me. That is for complaints. Just post right here.

Your question was answered in post #29, there is no known solution. Not every progressive dinner is possible, some just cannot be done. I have already given you some close solutions and you have accepted them then. Have you forgotten?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,848

Hi,

With this option,

(a) no couple serves at the same dinner more than once;

(b) no couple serves the same course more than once;

(c) each couple serves with either 8 or 9 different couples, and consequently there are either 11 or 10 couples (respectively) with whom they don't serve: eg,

(i) #1 serves with #s 2, 3, 4, 5, 9, 10, 14 and 18, but not with #s 6, 7, 8, 11, 12, 13, 15, 16, 17, 19 or 20;

(ii) #8 serves with #s 4, 5, 6, 7, 12, 13, 16, 17 and 20, but not with #s 1, 2, 3, 9, 10, 11, 14, 15, 18 or 19;

(d) each couple serves at least twice with 3 of the other couples (violating a rule in post #28, which was subsequently changed in post #36 to allow this): eg,

(i) #1 serves three times with #18, twice with #s 5 and 14, and once with #s 2, 3, 4, 9 and 10;

(ii) #8 serves twice with #s 4, 12 and 16, and once with #s 5, 6, 7, 13, 17 and 20.

```
| 1st course | 2nd course | 3rd course | 4th course | 5th course |
|-------------|-------------|-------------|-------------|-------------|
Dinner 1: | 1, 2, 3, 4 | 5, 6, 7, 8 | 9,10,11,12 | 13,14,15,16 | 17,18,19,20 |
Dinner 2: | 17, 5, 9,13 | 1,18,10,14 | 2,19, 6,15 | 3,20, 7,11 | 4, 8,12,16 |
Dinner 3: | 20, 8,12,16 | 4,17, 9,13 | 1,18, 5,14 | 2,19, 6,10 | 3, 7,11,15 |
Dinner 4: | 19, 7,11,15 | 3,20,12,16 | 4,17, 8,13 | 1,18, 5, 9 | 2, 6,10,14 |
```

*Last edited by phrontister (2014-10-10 18:24:47)*

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,786

Hi;

You will always have to violate one condition on this problem because I believe that it is impossible.

Hath ye seen the Sulemann challenge?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Olinguito****Member**- Registered: 2014-08-12
- Posts: 346

bobbym wrote:

The formal definition of the rational numbers is the field of fractions of the integral domain of the integers.
*Bassaricyon neblina*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,786

Hi;

I never said that. Please read post #1.

As I have said many times I am not a math guy, I am a computational guy. Thusly, I stay away from dry math definitions of the type best left to the theoreticians whom are experts in such things.

But since this is the kind of post JFF would deliver at me my heart has experienced a warm feeling. For that I thank you.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**ElainaVW****Member**- Registered: 2013-04-29
- Posts: 291

I think you misunderstood, the correction is of the post. It's not about you. It's not personal.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,786

When did I say or act that it was personal? I feel wonderful, so I post wonderfully. I thought my reply would brighten up Olinguito's day.

But by the way in answer to your bogus statement.

Everything is personal - Michael Corleone

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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