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

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klimenkovBest ResponseYou've already chosen the best response.1
Very interesting answer, because doesn't depend on \(n\).
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
is that a linear equation,\[x+y+z=b\] how many solutions
 one year ago

klimenkovBest ResponseYou've already chosen the best response.1
@experimentX actually you can think that N≤n, and there are no solutions if N>n. But the final formula will consider it.
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
dw:1350212084534:dw
 one year ago

asnaseerBest ResponseYou've already chosen the best response.0
if \(x_i\) represents ANY real number, then there are an infinite number of solutions
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
dw:1350212197575:dw
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
the problem is the difference between the solution of the equation and the solution set
 one year ago

klimenkovBest ResponseYou've already chosen the best response.1
@asnaseer I wrote in the statement of the problem that xi are natural.
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
dw:1350212336061:dw
 one year ago

asnaseerBest ResponseYou've already chosen the best response.0
ah! ok  thanks for pointing that out.
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
dw:1350180042769:dwslution set implies tha t there are infinite solutions to the numbers \[x _{1}+x _{2}=2\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
i only see 3 positive ones
 one year ago

klimenkovBest ResponseYou've already chosen the best response.1
@Jonask zero is not a natural number.
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
is it ?? \[ \binom {n+N1}{nN}\]
 one year ago

asnaseerBest ResponseYou've already chosen the best response.0
@klimenkov are you saying that you know what the answer is and that it does NOT depend on n?
 one year ago

klimenkovBest ResponseYou've already chosen the best response.1
@asnaseer that was reply on Jonask's answer.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.0
ok  that makes more sense now
 one year ago

klimenkovBest ResponseYou've already chosen the best response.1
@experimentX use your method to find the number of solutions for \(x_1=2\).
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
the easy element about this is that there are n constants
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
for n=N there should be 1 solutions. for n=N+1 there should be N solutions.
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
no constants (coeffecient)\[\left[\begin{matrix}1& ...1\\ & \end{matrix}\right]=\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
n = N+2, \[N + N(N1)/2\]
 one year ago

klimenkovBest ResponseYou've already chosen the best response.1
@experimentX for \(n=N+2\) that is not right.
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
huh .. isn't it \[ \binom{N}{1} + \binom{N}{2}\]
 one year ago

estudierBest ResponseYou've already chosen the best response.0
"there are no solutions if N>n" x_1 + x_2 = 3 (answer 1 +2) Am I missing something?
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
another un/lucky guess\[ \left(\begin{matrix}n1 \\ nN\end{matrix}\right) \]
 one year ago

klimenkovBest ResponseYou've already chosen the best response.1
@experimentX very close, but not right. Hope you will get it!
 one year ago

klimenkovBest ResponseYou've already chosen the best response.1
@estudier in your case N=2, n=3 N<n.
 one year ago

estudierBest ResponseYou've already chosen the best response.0
Ah, backwards ...duh!:)
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
@klimenkov what about n=N+2 case?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.0
After playing around with this for a while I /think/ the Fibonacci sequence is involved in the answer  or am I wildly off?
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
N=n >1 N=n1>1 N=n2>2 N=n3>3
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
N=1 >1 N=n/2 if n is even and (n1)/2 if n is odd
 one year ago

klimenkovBest ResponseYou've already chosen the best response.1
Wait a minute I think about your conclusions.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.0
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
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
That was only for n>5
 one year ago

estudierBest ResponseYou've already chosen the best response.0
Binomial coefficient, is it?
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
@experimentX and @estudier I think question is related to the necklace question... right?
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
\[ \binom{n1}{N1} = \left(\begin{matrix}N+21 \\ N1\end{matrix}\right) = \binom{N+1}{N1} = {(N+1)N \over 2} \] both seems to be equal here on this particular case. \[ \binom{N}{1} + \binom{N}{2} = N + {N(N1) \over 2} = {N(N+1) \over 2}\]
 one year ago

asnaseerBest ResponseYou've already chosen the best response.0
sorry, that should be:\[1+F(nN2)\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
@sauravshakya kind ov
 one year ago

klimenkovBest ResponseYou've already chosen the best response.1
Sorry guys, the are so many answers so I can't quickly answer. I must think about them. Especially about Fibonacci.
 one year ago

estudierBest ResponseYou've already chosen the best response.0
Stars and Bars ex 0 > (N1 n1)
 one year ago

asnaseerBest ResponseYou've already chosen the best response.0
Wait  I made a small mistake, I believe it should be:\[1+F(nN)\]And no problem @klimenkov  please take your time to check  this is a very interesting problem. :)
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
@klimenkov I know there must be a pattern but I dont think it is Fibonacci
 one year ago

klimenkovBest ResponseYou've already chosen the best response.1
I know the answer and it doesn't equal to Fibonacci number +1. Write the way you think @asnaseer
 one year ago

asnaseerBest ResponseYou've already chosen the best response.0
These were my thoughts: dw:1350214566420:dw
 one year ago

klimenkovBest ResponseYou've already chosen the best response.1
My English is not too good. But I think @estudier should be given Best response.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.0
Looking at this again, it looks more like:\[F(nN)+F(nN1)+...+F(1)\]
 one year ago

asnaseerBest ResponseYou've already chosen the best response.0
@estudier I did not understand what you meant by: Stars and Bars ex 0 > (N1 n1)
 one year ago

klimenkovBest ResponseYou've already chosen the best response.1
http://openstudy.com/users/klimenkov#/updates/50787c0ae4b02f109be43638 Try to get this.
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
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+n1]+where+n%3D3
 one year ago

estudierBest ResponseYou've already chosen the best response.0
Stars and Bars is combinatorics speak for putting N objects into n bins so that all bins contain at least one object.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.0
Oh I see  thanks for clarifying.
 one year ago

estudierBest ResponseYou've already chosen the best response.0
The idea of linking binomial coefficient with Fibonacci is interesting....
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
shouldn't this be like this (N1 n1) > (n1 N1) ??
 one year ago

klimenkovBest ResponseYou've already chosen the best response.1
@experimentX yes. The answer is \(\left(\begin{matrix}n1 \\ N1\end{matrix}\right)\). Who will prove it?
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
that and this are equivalent \[ \left(\begin{matrix}n1 \\ nN\end{matrix}\right) \]
 one year ago

estudierBest ResponseYou've already chosen the best response.0
Can we use k and n, I am getting very confused:)
 one year ago

klimenkovBest ResponseYou've already chosen the best response.1
Shame on me. Somebody clear my Smartscore...
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
we are just putting n=nN in your previous Q's answer ... whatever there is ... it is in your previous answer.
 one year ago

klimenkovBest ResponseYou've already chosen the best response.1
@experimentX was right since the very beginning.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.0
I don't understand how this can be correct. Take the n=5 and we will get:\[\frac{4!}{(4N)!(N1)!}\]then we get: N=1, Ans=4 N=2, Ans=12 Or have a misread the solution?
 one year ago

estudierBest ResponseYou've already chosen the best response.0
For positive n, k the number of different ntuples (ex 0) with sum k is (k1,n1)
 one year ago

klimenkovBest ResponseYou've already chosen the best response.1
@asnaseer \[\left(\begin{matrix}51 \\ N1\end{matrix}\right)=\frac{ 4! }{ (5N)!(N1)! }\]
 one year ago

asnaseerBest ResponseYou've already chosen the best response.0
D'oh! Now I think its time for someone to take my smartscore away! :)
 one year ago

klimenkovBest ResponseYou've already chosen the best response.1
Did anyone get this? Why does this formula give the right result?
 one year ago

estudierBest ResponseYou've already chosen the best response.0
Maybe an example helps to see it Seven identical coins between three people (k=7, n = 3) so (6,2) is 15 possible ways
 one year ago

estudierBest ResponseYou've already chosen the best response.0
There is a proof on page 12 of http://www.ma.utexas.edu/users/geir/teaching/m390c/lecturenotes.pdf
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
thanks estudier for sharing!!
 one year ago

asnaseerBest ResponseYou've already chosen the best response.0
I also extends my thanks to estudier. :)
 one year ago
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