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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

What table?

**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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

1;

1, 1, 1;

1, 2, 4, 6, 6;

1, 3, 9, 24, 54, 90, 90;

1, 4, 16, 60, 204, 600, 1440, 2520, 2520;

1, 5, 25, 120, 540, 2220, 8100, 25200, 63000, 113400, 113400;

1, 6, 36, 210, 1170, 6120, 29520, 128520, 491400, 1587600, 4082400,

It is in the examples in the link I posted.

The line

1, 5, 25, 120, 540, 2220, 8100, 25200, 63000, 113400, 113400;

is the one of interest.

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: 87,238

Hmmm, how do match that up with the pdf? Please demonstrate.

**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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

Hi bobbym

Do you see his table for P(X=n) on page 18?

Let T_n be the nth number in the 6th line of the triangle

1;

1, 1, 1;

1, 2, 4, 6, 6;

1, 3, 9, 24, 54, 90, 90;

1, 4, 16, 60, 204, 600, 1440, 2520, 2520;

1, 5, 25, 120, 540, 2220, 8100, 25200, 63000, 113400, 113400;

1, 6, 36, 210, 1170, 6120, 29520, 128520, 491400, 1587600, 4082400,

.

.

.

Then P(X=n)=T_(n-3)*Binomial[n-1,2]/6^(n-1)

*Last edited by anonimnystefy (2013-02-16 05:10:07)*

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: 87,238

Then what is i?

**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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

Sorry, wrong variable. Check again.

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: 87,238

Okay, you were supposed to check that but maybe you are not at your computer. I can check it. Supposing you are right is there an easier way to generate those numbers than he did?

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

I checked the for values of n from 3 to 9. I did not check the last 4.

I think it might be. I do not understand yet how the triangle was generated. I will have a look later.

Edit: OEIS gives a formula for those: T(k,n) = n![x^n](1+x+x^2/2)^k.

*Last edited by anonimnystefy (2013-02-16 04:43:26)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

Okay, I will need to work on it.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

For this particular problem the formula is T(n) = n![x^n](1+x+x^2/2)^5, which is kinda obvious, considering the fact that it is the number of ways we can put n marbles into 5 jars, with both jars being different and order of putting in mattering.

I think we actually solved it!

*Last edited by anonimnystefy (2013-02-16 04:50:56)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

Hold on! You solved it, but so far it is not working.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

Why not?

With the following code I get the correct answer!

`sum(i!/(2*6^(i-1))*coeff(ratexpand((1+x+x^2/2)^5),x^(i-3)),i,3,13),numer;`

*Last edited by anonimnystefy (2013-02-16 05:42:29)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

I was talking about the T[n] idea.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

It is working for me. What's the problem?

I think it could be generalized for 4, 5 and more dice in a row.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

Calm down please.

You are getting this to work?

*Last edited by bobbym (2013-02-16 05:06:58)*

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

I am calm. Sorry if I sounded frustrated.

T(n-3), not T(n), my bad. Edited the original post, too.

*Last edited by anonimnystefy (2013-02-16 05:10:24)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

I am kidding you. Just being funny.

I have it working now. Okay, you say you can improve on that formula?

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

I think it is plausible. Give me 5 minutes.

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: 87,238

I am working on it also.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

Hi bobbym

I am getting 11.21375706013234, or

in closed form.Edit: Here's my code:

`sum(i*(i-4)!/(6^(i-1))*binomial(i-1,3)*coeff(ratexpand((1+x+x^2/2+x^3/6)^5),x^(i-4)),i,4,19);`

*Last edited by anonimnystefy (2013-02-16 05:40:07)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

You are jumping ahead, what is that the answer to?

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

Expected number of throws till we get a number 4 times.

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: 87,238

Please hold on with that. We are still working on the 3. I agree that it will be possible to generalize but first I would like to simplify how you are doing the three. If we can do that we can really present it to him as a superior solution.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

Ok. So what do we do now?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

Is there a way to generate the sequence T[5,n] that does not require us to expand a trinomial or a sum with a ceiling function in it?

*Last edited by bobbym (2013-02-16 05:47:45)*

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

Combinatorics is Algebra and Algebra is Combinatorics.

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