Simple Combinatorics Problem

Agnishom · 2013-12-30 04:53:52

and then we collect the coefficient of x^20.

A post was deleted or I was daydreaming?

bobbym · 2013-12-30 04:55:55

Very, very good! How did you know that?

Two posts were deleted because I do not want to jump ahead with probability gf's just yet.

Now please compute the answer.

Agnishom · 2013-12-30 04:58:29

Looks like it is 35. How do I grep coefficients in Maxima? By the way, what does grepping mean?

bobbym · 2013-12-30 05:01:00

Grep is a command line utility for searching going back to Windows 98. Grepping must mean picking them out.

Agnishom · 2013-12-30 05:03:33

Is it a standard english word? How do I grep Coefficients in Maxima? Is 35 correct?

bobbym · 2013-12-30 05:06:01

Grepping is a technical term. 35 is correct and there should be a coefficient command in Maxima to do the job. But even if you can not find it we will not be doing ones that you can not just expand out.

Agnishom · 2013-12-30 05:09:22

But even if you can not find it we will not be doing ones that you can not just expand out.

What does it mean?

bobbym · 2013-12-30 05:11:49

What does what mean? Grepping or the quote?

Agnishom · 2013-12-30 05:15:21

Never mind; I've found the command

bobbym · 2013-12-30 05:19:01

Very good! I found it too.

How did you know to raise it up to the 4th power?

Agnishom · 2013-12-30 05:21:48

Ofcourse I cannot raise a sixterm polynomial to its 4th power by hand.

(%i12) coeff((expand((x + x^2 + x^3 + x^4 + x^5 + x^6)^4)),x,20);
(%o12)                                35

bobbym · 2013-12-30 05:24:01

Are you now ready to try your hand at a real combinatorics problem? One that would start to stump even the best NCR getters.

Agnishom · 2013-12-30 05:25:09

Yes; ...

bobbym · 2013-12-30 05:29:04

A mother wishes to share 20 dollars between Sally ,Edna, and
son Jack, so that they each receive at least $5, and nobody receives more than $10, and Edna gets an even amount. How many ways are there of doing this?

Agnishom · 2013-12-30 05:37:22

(x^5 + x^6 + x^7 + x^8 + x^9 + x^10)*(y^5 + y^6 + y^7 + y^8 + y^9 + y^10)*(z^5 + z^6 + z^7 + z^8 + z^9 + z^10)

???

bobbym · 2013-12-30 05:38:46

We do not need three different variables. I only did that for illustration purposes earlier.

Agnishom · 2013-12-30 05:47:19

If you do not need three different variables, then how can you ensure the fact about the even amount?

(%i18) coeff((expand(((x^5 + x^6 + x^7 + x^8 + x^9 + x^10)^2) * (x^6 + x^8 + x^10))),x,20);
(%o18)                                 9

xu drani

Last edited by Agnishom (2013-12-30 05:48:20)

bobbym · 2013-12-30 05:50:15

Excellent work! That is correct.

Supposing you were trying to convince me that gf's were better. Supposing you believed that, sum up in your own words why generating functions are easier.

Agnishom · 2013-12-30 05:52:55

Supposing you were trying to convince me that gf's were better. Supposing you believed that, sum up in your own words why generating functions are easier.

They are easier if you have a computer with you.

Agnishom · 2013-12-30 05:56:46

I still do not understand how you gAr got the gf for the original problem

bobbym · 2013-12-30 05:57:19

That is only the tip of the iceberg. Is

easy to do by hand?

Big problems usually generate big numbers, no matter what method you use you will need a computer. The real thing is that the gf approach reduces each problem to mere multiplication of polynomials! Once you formulate it and you see how easy that is even for a tough problem then you only have to multiply.

Agnishom · 2013-12-30 05:59:39

Agnishom wrote:

I still do not understand how you gAr got the gf for the original problem

bobbym · 2013-12-30 06:04:24

Do you understand this one?

Remember I said I only used the 3 variables for illustration. Well gAr, combined o and y into one variable.

And by the way, many of these can be expanded by the binomial theorem without a computer but not as quickly.

Agnishom · 2013-12-30 06:11:26

I do not understand why there is a 1 +

bobbym · 2013-12-30 06:15:27

(1 + o) means you can choose zero or one o.

Math Is Fun Forum

#26 2013-12-30 04:53:52

Re: Simple Combinatorics Problem

#27 2013-12-30 04:55:55

Re: Simple Combinatorics Problem

#28 2013-12-30 04:58:29

Re: Simple Combinatorics Problem

#29 2013-12-30 05:01:00

Re: Simple Combinatorics Problem

#30 2013-12-30 05:03:33

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#31 2013-12-30 05:06:01

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#32 2013-12-30 05:09:22

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#33 2013-12-30 05:11:49

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#34 2013-12-30 05:15:21

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#35 2013-12-30 05:19:01

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#36 2013-12-30 05:21:48

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#37 2013-12-30 05:24:01

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#38 2013-12-30 05:25:09

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#39 2013-12-30 05:29:04

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#40 2013-12-30 05:37:22

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#41 2013-12-30 05:38:46

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#42 2013-12-30 05:47:19

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