How many natural solutions does this equation have?
\[x_1+x_2+\ldots+x_N=n\]

- klimenkov

How many natural solutions does this equation have?
\[x_1+x_2+\ldots+x_N=n\]

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- schrodinger

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- anonymous

\[N\]

- experimentX

is N

- klimenkov

Very interesting answer, because doesn't depend on \(n\).

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## More answers

- anonymous

is that a linear equation,\[x+y+z=b\] how many solutions

- klimenkov

@experimentX actually you can think that N≤n, and there are no solutions if N>n. But the final formula will consider it.

- experimentX

|dw:1350212084534:dw|

- asnaseer

if \(x_i\) represents ANY real number, then there are an infinite number of solutions

- experimentX

|dw:1350212197575:dw|

- anonymous

the problem is the difference between the solution of the equation and the solution set

- klimenkov

@asnaseer I wrote in the statement of the problem that xi are natural.

- experimentX

|dw:1350212336061:dw|

- asnaseer

ah! ok - thanks for pointing that out.

- anonymous

|dw:1350180042769:dw|slution set implies tha t there are infinite solutions to the numbers
\[x _{1}+x _{2}=2\]

- anonymous

(0,2)
(2,0)\
(1,1)

- anonymous

i only see 3 positive ones

- klimenkov

@Jonask zero is not a natural number.

- experimentX

is it ??
\[ \binom {n+N-1}{n-N}\]

- asnaseer

@klimenkov are you saying that you know what the answer is and that it does NOT depend on n?

- klimenkov

@asnaseer that was reply on Jonask's answer.

- asnaseer

ok - that makes more sense now

- klimenkov

@experimentX use your method to find the number of solutions for \(x_1=2\).

- anonymous

the easy element about this is that there are n constants

- experimentX

for n=N there should be 1 solutions.
for n=N+1 there should be N solutions.

- anonymous

no constants (coeffecient)\[\left[\begin{matrix}1& ...1\\ & \end{matrix}\right]=\]

- anonymous

the coeffecient matrix

- experimentX

n = N+2,
\[N + N(N-1)/2\]

- klimenkov

@experimentX for \(n=N+2\) that is not right.

- experimentX

huh .. isn't it
\[ \binom{N}{1} + \binom{N}{2}\]

- anonymous

"there are no solutions if N>n"
x_1 + x_2 = 3 (answer 1 +2) Am I missing something?

- experimentX

another un/lucky guess\[
\left(\begin{matrix}n-1 \\ n-N\end{matrix}\right)
\]

- klimenkov

@experimentX very close, but not right. Hope you will get it!

- klimenkov

@estudier in your case N=2, n=3 N

- anonymous

Ah, backwards ...duh!:-)

- experimentX

@klimenkov what about n=N+2 case?

- experimentX

is that correct?

- asnaseer

After playing around with this for a while I /think/ the Fibonacci sequence is involved in the answer - or am I wildly off?

- anonymous

N=n --->1
N=n-1-->1
N=n-2--->2
N=n-3--->3

- anonymous

N=1 --->1
N=n/2 if n is even and (n-1)/2 if n is odd

- klimenkov

Wait a minute I think about your conclusions.

- asnaseer

I am getting something along the lines of:\[1+F(x)\]where x is some function of n and N and F(x) is the x'th Fibonacci number

- anonymous

That was only for n>5

- anonymous

Binomial coefficient, is it?

- anonymous

@experimentX and @estudier I think question is related to the necklace question... right?

- asnaseer

is it:\[1+F(N-n-2)\]

- anonymous

N-1,n-1

- experimentX

\[ \binom{n-1}{N-1} = \left(\begin{matrix}N+2-1 \\ N-1\end{matrix}\right) = \binom{N+1}{N-1} = {(N+1)N \over 2} \]
both seems to be equal here on this particular case.
\[ \binom{N}{1} + \binom{N}{2} = N + {N(N-1) \over 2} = {N(N+1) \over 2}\]

- asnaseer

sorry, that should be:\[1+F(n-N-2)\]

- experimentX

@sauravshakya kind ov

- asnaseer

that is for n>N+2

- klimenkov

Sorry guys, the are so many answers so I can't quickly answer. I must think about them. Especially about Fibonacci.

- anonymous

Stars and Bars ex 0 -> (N-1 n-1)

- asnaseer

Wait - I made a small mistake, I believe it should be:\[1+F(n-N)\]And no problem @klimenkov - please take your time to check - this is a very interesting problem. :)

- anonymous

@klimenkov I know there must be a pattern but I dont think it is Fibonacci

- klimenkov

I know the answer and it doesn't equal to Fibonacci number +1. Write the way you think @asnaseer

- asnaseer

These were my thoughts:
|dw:1350214566420:dw|

- klimenkov

My English is not too good. But I think @estudier should be given Best response.

- asnaseer

Looking at this again, it looks more like:\[F(n-N)+F(n-N-1)+...+F(1)\]

- asnaseer

@estudier I did not understand what you meant by:
Stars and Bars ex 0 -> (N-1 n-1)

- klimenkov

- experimentX

for n=6 and N=3, the answer is 10 ... this seems to work out
http://www.wolframalpha.com/input/?i=Binomial[n%2B2%2C+n-1]+where+n%3D3

- anonymous

Stars and Bars is combinatorics speak for putting N objects into n bins so that all bins contain at least one object.

- asnaseer

Oh I see - thanks for clarifying.

- anonymous

The idea of linking binomial coefficient with Fibonacci is interesting....

- experimentX

shouldn't this be like this
(N-1 n-1) -> (n-1 N-1) ??

- klimenkov

@experimentX yes. The answer is \(\left(\begin{matrix}n-1 \\ N-1\end{matrix}\right)\). Who will prove it?

- experimentX

that and this are equivalent
\[ \left(\begin{matrix}n-1 \\ n-N\end{matrix}\right) \]

- anonymous

Can we use k and n, I am getting very confused:-)

- klimenkov

Shame on me. Somebody clear my Smartscore...

- asnaseer

:)

- experimentX

we are just putting n=n-N in your previous Q's answer ... whatever there is ... it is in your previous answer.

- klimenkov

@experimentX was right since the very beginning.

- asnaseer

I don't understand how this can be correct. Take the n=5 and we will get:\[\frac{4!}{(4-N)!(N-1)!}\]then we get:
N=1, Ans=4
N=2, Ans=12
Or have a misread the solution?

- anonymous

For positive n, k the number of different n-tuples (ex 0) with sum k is (k-1,n-1)

- klimenkov

@asnaseer
\[\left(\begin{matrix}5-1 \\ N-1\end{matrix}\right)=\frac{ 4! }{ (5-N)!(N-1)! }\]

- asnaseer

D'oh! Now I think its time for someone to take my smartscore away! :)

- klimenkov

Did anyone get this? Why does this formula give the right result?

- anonymous

Maybe an example helps to see it
Seven identical coins between three people (k=7, n = 3) so (6,2) is 15 possible ways

- anonymous

There is a proof on page 12 of
http://www.ma.utexas.edu/users/geir/teaching/m390c/lecturenotes.pdf

- experimentX

thanks estudier for sharing!!

- anonymous

yw:-)

- asnaseer

I also extends my thanks to estudier. :)

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