number of solutions to a linear equation

gourish · 2014-12-14 05:31:57

umm... take x_xxxxxx_xxx_xxx.... so there are groups (x) (xxxxxx) (xxx) (xxx).... let a=(x) b=(xxxxxx) c=(xxx) d=(xxx)..

the sum of these groups or variables is 13. and i took the c and d as the same as there is no constraint like a,b,c,d are distinct....

bobbym · 2014-12-14 05:34:41

Okay, so for your problem we only need to know in the nine spaces

between x x x x x x x x x x how many ways there are to put 2 dashes.

gourish · 2014-12-14 05:39:08

well that's just choosing 2 spaces out of 9.... cool i got it it's (9 chooses 2)... brilliant...

gourish · 2014-12-14 05:40:51

but there's this one problem how did u find that there are 9 spaces when u know that there are 10 x's.... you can put a dash in place of each x.. so there must be 10 spaces right? what am i missing here....?

gourish · 2014-12-14 05:42:26

oh wait... i can't... i got it... i can not put a dash at an end... that would mean that one of the groups (variable) is zero which is against my constraint...

Agnishom · 2014-12-14 05:43:18

You have to choose two of these dashes:
X_x_x_x_x_x_x_x_x_x

bobbym · 2014-12-14 05:43:34

The generating function does the calculating without doing any of the above reasoning.

x can be any number from 1 to 10 so it is represented by the polynomial

y can be any number from 1 to 10 so it is represented by the polynomial

z can be any number from 1 to 10 so it is represented by the polynomial

gourish · 2014-12-14 05:47:29

x can be any number from one to ten but why does it mean that it can be represented as x+x^2 and so on till x^10 ?

bobbym · 2014-12-14 05:55:49

That is what they call placeholders. The powers of each x represents the number. It is a little complicated and a book is the best place to get a detailed explanation. Anyway, they do the combinatorics for us.

gourish · 2014-12-14 05:56:38

okay.... go on... i have the polynomials representing the variables...

Agnishom · 2014-12-14 06:00:29

We want to add x, y, z to get 10. So, to add their exponents we multiply the polynomials.

So, we need to multiply

(x+x^2+x^3+...)(x+x^2+x^3+...)(x+x^2+x^3+...)

bobbym · 2014-12-14 06:05:14

This is pretty mechanical so you just use the same process basically again and again.

You just multiply the three polynomials:

this equals when expanded.

You now check the coefficient of x^10, what is it?

gourish · 2014-12-14 06:05:27

okay so we now get to back the question that u originally asked me.... right... we now need to find the coefficient of x^10 in this expansion where the whole term is raised to 3....

now getting back to the multinomial theorem....(i have got a fair enough idea of binomial theorem and have read that multinomial theorem is general form of bi theo..)

bobbym · 2014-12-14 06:06:41

One thing at a time.

This is pretty mechanical so you just use the same process basically again and again.

You just multiply the three polynomials:

this equals when expanded.

You now check the coefficient of x^10, what is it?

gourish · 2014-12-14 06:07:59

36... the answer...

i bet you didn't expand that term by hand....

Agnishom · 2014-12-14 06:08:27

We used a computer!

gourish · 2014-12-14 06:10:00

so any suggestions for a good book that i can read for combinatorics... including such problems...

Agnishom · 2014-12-14 06:11:13

The tucker book!

gourish · 2014-12-14 06:13:40

that's it's name? "tucker book" on google gave me links to amazon having weird books "they serve beer in hell" can you be more specific....

bobbym · 2014-12-14 06:14:49

i bet you didn't expand that term by hand....

It can be done by hand too and books are full of ways to show you how.

Try this book Applied Combinatorics sixth edition Alan Tucker.

gourish · 2014-12-14 06:17:54

well does it cover the basic combinatorics as well... i think i will need to rebuild my basics to be able to fully appreciate it's applications

Agnishom · 2014-12-14 06:20:24

Yes, it provides a very strong base for combinatorial reasoning. That is better than books with around 100-200 problems with just 4-5 line solutions without any explanation

bobbym · 2014-12-14 06:21:09

It covers everything but it is tough. Do not read it as you were taught to read a textbook.

gourish · 2014-12-14 06:22:37

ok... well i read that it contains graph theory... why is that necessary for combinatorics?

bobbym · 2014-12-14 06:27:26

I do not use it so much myself but it is called by some the replacement for topology or rather the poor man's topology. You have just seen how a combinatorics problem can be turned into an analysis problem when we use generating functions. The ability to change a problem from one branch of math to another is the key to solving more problems.

Math Is Fun Forum

#26 2014-12-14 05:31:57

Re: number of solutions to a linear equation

#27 2014-12-14 05:34:41

Re: number of solutions to a linear equation

#28 2014-12-14 05:39:08

Re: number of solutions to a linear equation

#29 2014-12-14 05:40:51

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#30 2014-12-14 05:42:26

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#31 2014-12-14 05:43:18

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#32 2014-12-14 05:43:34

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#33 2014-12-14 05:47:29

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#34 2014-12-14 05:55:49

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#35 2014-12-14 05:56:38

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#36 2014-12-14 06:00:29

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#37 2014-12-14 06:05:14

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#38 2014-12-14 06:05:27

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#39 2014-12-14 06:06:41

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#40 2014-12-14 06:07:59

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#41 2014-12-14 06:08:27

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#42 2014-12-14 06:10:00

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#43 2014-12-14 06:11:13

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#44 2014-12-14 06:13:40

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#45 2014-12-14 06:14:49

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#46 2014-12-14 06:17:54

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#47 2014-12-14 06:20:24

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#48 2014-12-14 06:21:09

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#49 2014-12-14 06:22:37

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