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

Mathematics
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\[N\]
is N
Very interesting answer, because doesn't depend on \(n\).

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Other answers:

is that a linear equation,\[x+y+z=b\] how many solutions
@experimentX actually you can think that N≤n, and there are no solutions if N>n. But the final formula will consider it.
|dw:1350212084534:dw|
if \(x_i\) represents ANY real number, then there are an infinite number of solutions
|dw:1350212197575:dw|
the problem is the difference between the solution of the equation and the solution set
@asnaseer I wrote in the statement of the problem that xi are natural.
|dw:1350212336061:dw|
ah! ok - thanks for pointing that out.
|dw:1350180042769:dw|slution set implies tha t there are infinite solutions to the numbers \[x _{1}+x _{2}=2\]
(0,2) (2,0)\ (1,1)
i only see 3 positive ones
@Jonask zero is not a natural number.
is it ?? \[ \binom {n+N-1}{n-N}\]
@klimenkov are you saying that you know what the answer is and that it does NOT depend on n?
@asnaseer that was reply on Jonask's answer.
ok - that makes more sense now
@experimentX use your method to find the number of solutions for \(x_1=2\).
the easy element about this is that there are n constants
for n=N there should be 1 solutions. for n=N+1 there should be N solutions.
no constants (coeffecient)\[\left[\begin{matrix}1& ...1\\ & \end{matrix}\right]=\]
the coeffecient matrix
n = N+2, \[N + N(N-1)/2\]
@experimentX for \(n=N+2\) that is not right.
huh .. isn't it \[ \binom{N}{1} + \binom{N}{2}\]
"there are no solutions if N>n" x_1 + x_2 = 3 (answer 1 +2) Am I missing something?
another un/lucky guess\[ \left(\begin{matrix}n-1 \\ n-N\end{matrix}\right) \]
@experimentX very close, but not right. Hope you will get it!
@estudier in your case N=2, n=3 N
Ah, backwards ...duh!:-)
@klimenkov what about n=N+2 case?
is that correct?
After playing around with this for a while I /think/ the Fibonacci sequence is involved in the answer - or am I wildly off?
N=n --->1 N=n-1-->1 N=n-2--->2 N=n-3--->3
N=1 --->1 N=n/2 if n is even and (n-1)/2 if n is odd
Wait a minute I think about your conclusions.
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
That was only for n>5
Binomial coefficient, is it?
@experimentX and @estudier I think question is related to the necklace question... right?
is it:\[1+F(N-n-2)\]
N-1,n-1
\[ \binom{n-1}{N-1} = \left(\begin{matrix}N+2-1 \\ N-1\end{matrix}\right) = \binom{N+1}{N-1} = {(N+1)N \over 2} \] both seems to be equal here on this particular case. \[ \binom{N}{1} + \binom{N}{2} = N + {N(N-1) \over 2} = {N(N+1) \over 2}\]
sorry, that should be:\[1+F(n-N-2)\]
that is for n>N+2
Sorry guys, the are so many answers so I can't quickly answer. I must think about them. Especially about Fibonacci.
Stars and Bars ex 0 -> (N-1 n-1)
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. :)
@klimenkov I know there must be a pattern but I dont think it is Fibonacci
I know the answer and it doesn't equal to Fibonacci number +1. Write the way you think @asnaseer
These were my thoughts: |dw:1350214566420:dw|
My English is not too good. But I think @estudier should be given Best response.
Looking at this again, it looks more like:\[F(n-N)+F(n-N-1)+...+F(1)\]
@estudier I did not understand what you meant by: Stars and Bars ex 0 -> (N-1 n-1)
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
Stars and Bars is combinatorics speak for putting N objects into n bins so that all bins contain at least one object.
Oh I see - thanks for clarifying.
The idea of linking binomial coefficient with Fibonacci is interesting....
shouldn't this be like this (N-1 n-1) -> (n-1 N-1) ??
@experimentX yes. The answer is \(\left(\begin{matrix}n-1 \\ N-1\end{matrix}\right)\). Who will prove it?
that and this are equivalent \[ \left(\begin{matrix}n-1 \\ n-N\end{matrix}\right) \]
Can we use k and n, I am getting very confused:-)
Shame on me. Somebody clear my Smartscore...
:)
we are just putting n=n-N in your previous Q's answer ... whatever there is ... it is in your previous answer.
@experimentX was right since the very beginning.
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?
For positive n, k the number of different n-tuples (ex 0) with sum k is (k-1,n-1)
@asnaseer \[\left(\begin{matrix}5-1 \\ N-1\end{matrix}\right)=\frac{ 4! }{ (5-N)!(N-1)! }\]
D'oh! Now I think its time for someone to take my smartscore away! :)
Did anyone get this? Why does this formula give the right result?
Maybe an example helps to see it Seven identical coins between three people (k=7, n = 3) so (6,2) is 15 possible ways
There is a proof on page 12 of http://www.ma.utexas.edu/users/geir/teaching/m390c/lecturenotes.pdf
thanks estudier for sharing!!
yw:-)
I also extends my thanks to estudier. :)

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