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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,592

Hi Bobby;

Have you had a look yet at my post #24 answer to the 13-athletes question from post #1? Just wondering if you agree with my 270270, or have something else.

"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****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

I have not had any time to look at it. I have been setting up a new machine and it has been a real bear.

The problem with your answer is this: we only have that answer and an experimental answer like that requires us to get the answer in two different ways. Once we have agreement in the two different methods we have an extremely high probability that we are correct. Calculating how high...that too is very difficult.

Until this machine stops fighting with me we will accept your answer. Now, I answer this a couple of ganesh's and eat and back to the setting up. Hopefully, I will be able to verify your answer soon. In the meantime, can I see what you did?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,592

Hi Bobby;

In the meantime, can I see what you did?

In post #24 I explained what I did (although maybe not fully), but I didn't show my M code that I adapted from yours. This is it:

```
n = 11;
s = Permutations[Range[n]];
teams = Partition[#, 2] & /@ s;
DeleteDuplicates[Map[Sort, Most /@ teams, 2]] // Length
```

It's only good for n=an odd number (your code does even numbers), for which n=11 (answer 17325) is the highest my computer will go...because of the 'combinatorial explosion'.

The following code is from OEIS. I changed the min and max and divided by 2 to suit our problem.

`Table[(2 n + 5)!!/3 - (2 n + 3)!!, {n, 1, 5}]/2`

The adapted code outputs all five levels, the last being the 270270.

My images in post #24 hopefully will help show how I identified the pattern.

*Last edited by phrontister (2017-02-15 16:45:06)*

"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****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

Your answer generates the sequence:

10, 105, 1260, 17325, 270270,... is the 10 correct for 1 pair?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,592

I believe it is correct.

In the figures below, which are from level 1, the first column lists the 10 valid pairs that can be formed from the five athletes {1,2,3,4,5} on the roster, subject to the constraint that there are 3 more athletes on the roster than the number chosen to play (same as for level 5). So the second column lists those who miss out.

For the STP I had to decide how, for levels 1-4, I should deal with the difference of 3 that exists in level 5 (ie, 13-10), and I opted for maintaining it consistently throughout all levels.

That is why all 10 lines contain just one chosen pair, and a threesome who missed out.

```
1,2 3,4,5
1,3 2,4,5
1,4 2,3,5
1,5 2,3,4
2,3 1,4,5
2,4 1,3,5
2,5 1,3,4
3,4 1,2,5
3,5 1,2,4
4,5 1,3,2
```

But my thinking may be wrong on that, so please convince me otherwise if I've erred.

"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****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

I have not been able to run a full count of this problem even with my new machine buzzing away like a bee. So, I do not know the correct answer at this time.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,592

I sure hope that the buzzing sound in your new machine is not from a sick bee, or something more sinister!

Must go to bed now...see you tomorrow.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Okay, I will post when I have something. Pleasant dreams.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Beach Coach****Member**- Registered: 2017-02-09
- Posts: 5

I'm learning from watching this process!

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,592

It's nice to know that you're on board, Coach!

Often an OP disappears after posting their problem, despite the fact that a full-on discussion by forum members about its solution may follow!

Hopefully we can get somewhere with yours...

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**Beach Coach****Member**- Registered: 2017-02-09
- Posts: 5

I'm following it closely. There are a lot of coaches who will be very interested in these numbers. Beach volleyball is the fastest growing NCAA sport in history. It is also growing in the So Cal high schools and in other states also, and as a club sport.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,592

No pressure on us, then!

I'm enjoying the struggle!

*Last edited by phrontister (2017-02-16 11:46:37)*

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi phrontister;

Let us do one together to see if methods and results check. What do you get for 3 pairs from a group of 9 players?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,592

Hi Bobby;

What do you get for 3 pairs from a group of 9 players?

I get 1260 possible combinations of 3 chosen pairs and their accompanying threesomes who are omitted. That totals 9 athletes for each combination.

See my image in the 'level 3' hidebox in post #24. It lists the first and last 60 combinations that I got from my M code. The chosen player groups are highlighted in yellow.

```
n = 9;
s = Permutations[Range[n]];
teams = Partition[#, 2] & /@ s;
DeleteDuplicates[Map[Sort, Most /@ teams, 2]];
Length[%]
%%
```

*Last edited by phrontister (2017-02-16 19:26:50)*

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

I agree! How about 2 pairs out of 7 players?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,592

I get 105 possible combinations of 2 chosen pairs and their accompanying threesomes who are omitted.

My image in the 'levels 1&2' hidebox in post #24 covers that one (players in orange)...and also the one for 5 athletes (players in blue).

In my images, the combination totals and the number of athletes involved are shown at the bottom of the combination lists. Each list has its own colour, highlighting the chosen players.

*Last edited by phrontister (2017-02-16 19:37:35)*

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

I agree with 105. How about 1 more? Can you calculate 4 pairs out of 9 people?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**iamaditya****Member**- From: Planet Mars
- Registered: 2016-11-15
- Posts: 692

iamaditya wrote:

Hi;

After getting more details i found that the total no. Of possibilities are 10C2 or C(10,2) =10!/(2!*8!)=45

Practice makes a man perfect.

There is no substitute to hard work

All of us do not have equal talents but everybody has equal oppurtunities to build their talents.-APJ Abdul Kalam

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,592

bobbym wrote:

Can you calculate 4 pairs out of 9 people?

That's 17325 possible combinations of 4 chosen pairs and their accompanying threesomes who are omitted.

My image in the 'level 4' hidebox in post #24 covers that one (players in green).

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

Can you calculate 4 pairs out of 9 people?

I do not get that many. I think the difference in our answers is because you solved for 11 people.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,592

Oops, so I did. I'll try again!

*Last edited by phrontister (2017-02-17 01:40:55)*

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,592

945

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

I agree. I am using a different method than you are so we can be pretty sure of those answers.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,592

Excellent!

Does your code solve for 5 pairs out of 13 athletes?

My computer can't handle that number of permutations. M stops working straightaway, giving this message:

"General::nomem: The current computation was aborted because there was insufficient memory available to complete the computation."

I noticed that when I ran the M code for 11 athletes, it used up all available RAM (20GB or so) and then began to use virtual memory from the hard drive...which turned out to be enough to get a result. The extra amount of memory needed to solve for 13 athletes would be absolutely humongous, so I let that one slide and just took OEIS's word for it that 270270 was the next element after 17325 in the sequence.

Their formula, (2n+3)!/(3!*n!*2^n), and their M code, Table[(2n+5)!!/3-(2n+3)!!,{n,1,5}]/2, confirm that. Of course, I don't understand the how/what/why etc of those formulas!

*Last edited by phrontister (2017-02-17 03:37:22)*

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

Could you please run this when you get a chance and post a screenshot of the notebook that pops up.

```
Needs["Benchmarking`"]
BenchmarkReport[]
```

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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