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**zetafunc.****Guest**

I will have to learn to adapt.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,260

Once they begin to see you in one of the 9 or 10 forms that they can not do without, then you will do better.

**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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**zetafunc.****Guest**

Until then, it is a waiting game...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,260

I suggest doing your math and forgetting about them. Not completely but putting them in the last and smallest compartment of your mind.

**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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**zetafunc.****Guest**

I find it difficult to forget about them...

I finally found a STEP question that was easily done with generating functions, getting the co-efficient was just a bit tedious.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,260

The tough part of any generating function approach is getting the coefficient.

**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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**zetafunc.****Guest**

Yes, unfortunately no calc/CAS is allowed for STEP. Luckily all I had to do was get the coefficient of x[sup]24[/sup] in (1 + x + x[sup]2[/sup] + x[sup]3[/sup] + x[sup]4[/sup] + x[sup]5[/sup] + x[sup]6[/sup])[sup]5[/sup]. If it was an EGF that would have been more tedious.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,260

Hi;

I was working on a paper for just that sort of problem. I will back soon from doing a chore.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

A paper?

Okay, see you later.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,260

Only thing it did not have the constant term.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Sorry, I had to go to school, I am back now.

So you have a good way of getting the co-efficient?

My method was

and then just get the co-efficient of x^24 from that but then the arithmetic can become a bit long...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,260

You can play spot the pattern.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Okay............

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,260

Supposing you needed the coefficient of x^5 in

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

By hand, I think I would just go back to sleep if someone asked me to do that

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,260

That could be difficult even by computer. That is when math becomes an art and ceases to be a science.

Anyway this is the answer:

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

How do you know that?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,260

Sometimes in combinatorics we have a big problem. We can often solve it by looking at smaller problems that have the same flavor and then spotting the pattern. For instance it is not too difficult to expand these:

What do you notice about the coefficient of x^5 in each case?

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

The co-efficients are constant for the first sequence, (n+1) for the second sequence, and the triangular numbers for the third.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,260

You can put that into a more succinct combinatoric notation.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

(n+1) choose (k-1), where k is the power of the big bracket?

**zetafunc.****Guest**

Hmm maybe (n + k - 1) choose (k - 1).

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,260

You can phrase wholly in terms of the power n:

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Oh I see. So in my example I could have just done

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,260

Yes, because of symmetry:

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

Combinatorics is Algebra and Algebra is Combinatorics.

**Online**