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**Beach Coach****Guest**

I coach beach volleyball. We field 5 pairs of athletes. How many possible combinations of 5 pairs I can make from 10 athletes?

And if you're able to answer that one, here's a further question. We actually have 13 athletes on the roster. 10 of them can play in a match (5 doubles pairs). How many possible combinations of 5 pairs can I make from 13 athletes?

Thank you!!

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

Hi;

I coach beach volleyball. We field 5 pairs of athletes. How many possible combinations of 5 pairs I can make from 10 athletes?

**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: 725

Hi;

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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**Mathegocart****Member**- Registered: 2012-04-29
- Posts: 1,882

bobbym wrote:

Hi;

I coach beach volleyball. We field 5 pairs of athletes. How many possible combinations of 5 pairs I can make from 10 athletes?

Another GF?

The integral of hope is reality.

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

Nope, this one was entirely done by a very simple tool used in EM. It is called "Spot the Pattern."

**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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**Mathegocart****Member**- Registered: 2012-04-29
- Posts: 1,882

wait a moment... i see a pattern!

The integral of hope is reality.

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

**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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**Mathegocart****Member**- Registered: 2012-04-29
- Posts: 1,882

*Last edited by Mathegocart (2017-02-08 14:35:00)*

The integral of hope is reality.

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

Hi;

**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

Here's more detail:

There are 10 athletes.

We make 5 pairs to play in matches.

Any of the 10 athletes can be paired with any of the other 9 to make a pair.

There are always 5 pair.

So far, the answers posted are 45, 250 and 945.

Having made these pairs for 10 seasons, I've realized there are a lot of possible pairings. I was thinking it would be 5,000 to 6,500 or around 25,000. I'm a Harvard grad but not a math whiz, so I appreciate your input greatly.

If you'll share your formula so all who are watching can help, that would be great.

Thank you to all of you who have helped so far.

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

Hi coach;

I am sure somebody can come up with a formula that will spit out 25000 or any other number you want but until someone shows me a program that counts more than 945 ways, I am going to stick with that.

**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;

I got 945 (as Bobby did), also by playing "Spot the Pattern".

OEIS gives that answer too, and the formula (2*n)!/(n!*2^n). In our case *n* is the number of pairs of athletes.

The answer is the product of the first *n* odd numbers: 1*3*5*7*9 = 945.

I did the first four elements of the pattern by hand, but stopped there to play STP as the fifth was going to mean I'd be writing into the middle of next week!

"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;

Yes! STP! And for backup...OEIS! Wunderbar!

**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

How else!

Here are images of my work for the STP:

*Last edited by phrontister (2017-02-23 00:01:08)*

"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

How else!

Did you try a bit of M?

**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

Yes I did, but didn't get there.

I've been looking at Subsets[Range], and also DeleteDuplicates[Sort/@Permutations[Range[]]]. Still looking at it...

*Last edited by phrontister (2017-02-12 02:40:32)*

"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

It is a bit tricky.

**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'll say!

Still looking...

Any hints?

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

Yes, when you get the answer save the notebook! I deleted mine and now can not duplicate my work.

**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

Not quite the kind of hint I was looking for, but good advice!

Maybe a data recovery program could find the deleted file...if it hasn't become too scrambled over time.

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

Never had much luck with those.

Try this:

```
n = 1;
s = Permutations[Range[2 n]];
teams = Partition[#, 2] & /@ s;
DeleteDuplicates[Map[Sort, teams, 2]] // Length
```

Adjust n for higher values. Suffers from combinatorial explosion but is short.

**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

Yes, that's it! I've saved it, so if you ever need a copy...!

Not being able to make pairs had been holding me up, but I hadn't thought of using Partition. Never used it before.

I haven't tried problem #2 (the one about 13 athletes) in the first post yet, but I'm off to bedzzzzzz....

Catch you later.

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

Hi;

Okay, see you later.

**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

Beach Coach wrote:

We actually have 13 athletes on the roster. 10 of them can play in a match (5 doubles pairs). How many possible combinations of 5 pairs can I make from 13 athletes?

270270, I think.

I played STP by hand in Excel for levels 1 & 2, but the size of the problem was escalating enormously and I had to change tack. So then I adapted Bobby's M code from post #21 to fit my levels 1 & 2 results and got levels 3 & 4 (but then ran out of computer memory, preventing further progress that way).

At that stage I had the sequence {10,105,1260,17325}, covering the first 4 levels.

At OEIS I found the last (5th) level and the following formula for the sequence:

(2n+3)!/(3!*n!*2^n)

For our problem, *n* is the number of doubles pairs (5), from which (2*5+3)!/(3!*5!*2^5) = 270270.

Here are images of my work for the STP:

*Last edited by phrontister (2017-02-23 00:02:44)*

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

First of all, THANK YOU for all this great input and thinking!

So, is this the consensus:

ANSWER: (2*n)!/(n!*2^n). In our case n is the number of pairs of athletes. The answer is the product of the first n odd numbers: 1*3*5*7*9 = 945.

Here's a further question. We actually have 13 athletes on the roster. 10 of them can play in a match (5 doubles pairs). How many possible combinations of 5 pairs can I make from 13 athletes?

ANSWER: (2n+3)!/(3!*n!*2^n) - For our problem, n is the number of doubles pairs (5), from which (2*5+3)!/(3!*5!*2^5) = 270270.

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