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**Dab Alex****Guest**

I have a problem I need to solve so I'd need an answer or a way to solve it.

I need to find out the number of combinations of 6 numbers from 49 numbers. The tricky part (for me )is that I want to exclude the combinations of numbers that are in sequence of minimum 3 meaning:

I want to exclude results like {1,2,3,4,5,6} or like {1,2,3,9,12,14} or {1,2,3,4,6,8} so on....

Thank you!

**Dab Alex****Guest**

I forgot to mention.... without repetitions and order is not important

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

Hi;

If order does not count then doesn't {2,46,31,1,5,3} have to excluded? There is a {1,2,3} in there.

*Last edited by bobbym (2013-02-19 14:53:23)*

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**Dab Alex****Guest**

I mean that the sequence of consecutive numbers must not be contained in the result so this result {2,46,31,1,5,3} must also be excluded.

I specified that order is not important because no. of combinations if order is important is 1.00683475e+10 and if it isn't no. of combinations is 13983816

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

Hi;

Yes, you are correct there are 10068347520 permutations and only 13983816 combinations.

I think I have enough to start working on it, thanks.

I am getting that 13316842 out of 13983816 do not have 3 or more numbers in sequence. This agrees well with simulations.

*Last edited by bobbym (2013-02-21 04:50:49)*

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

I think the reccurence is:

A(n,k)=A(n-3,k-2)+A(n-2,k-1)+A(n-1,k), n>3, with the appropriate starting conditions.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

Hmmmm? Are you sure?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

Preety much.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

What are your initial conditions?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

A(n,k) is 0 for n<=0, except A(0,0)=1.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

You also did not define n or k. What are they?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

Numbers, of course.

Joke aside, n is how many numbers we choose from and k is how many numbers we choose.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

k is how many numbers we choose.

You are saying n=6 and k is the number of consecutives?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

No. n=49 and k=6.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

Did you check it by actually running the recurrence? Does it get my answer?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

Hi bobbym

Sorry. Take A(0,1)=1 as well.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

Hi;

The same question as the last post. Did you try it?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

I'm calculating it. It will take a while.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

That is like music to my ears!

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

Do you think there is a way to speed it up?

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

I do not know what you are doing, so how can I say.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

I am just running the recurrence.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

That is what I mean. In Maxima?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

Yes.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

Well, when I ask this question the kaboobly doo will hit the fan. Where does it come from?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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