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klimenkov
Group Title
How many natural solutions does this equation have?
\[x_1+x_2+\ldots+x_N=n\]
 2 years ago
 2 years ago
klimenkov Group Title
How many natural solutions does this equation have? \[x_1+x_2+\ldots+x_N=n\]
 2 years ago
 2 years ago

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experimentX Group TitleBest ResponseYou've already chosen the best response.3
is N<n ??
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
Very interesting answer, because doesn't depend on \(n\).
 2 years ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
is that a linear equation,\[x+y+z=b\] how many solutions
 2 years ago

klimenkov Group TitleBest 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.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
dw:1350212084534:dw
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.0
if \(x_i\) represents ANY real number, then there are an infinite number of solutions
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
dw:1350212197575:dw
 2 years ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
the problem is the difference between the solution of the equation and the solution set
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
@asnaseer I wrote in the statement of the problem that xi are natural.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
dw:1350212336061:dw
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.0
ah! ok  thanks for pointing that out.
 2 years ago

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

Jonask Group TitleBest ResponseYou've already chosen the best response.0
(0,2) (2,0)\ (1,1)
 2 years ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
i only see 3 positive ones
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
@Jonask zero is not a natural number.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
is it ?? \[ \binom {n+N1}{nN}\]
 2 years ago

asnaseer Group TitleBest 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?
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
@asnaseer that was reply on Jonask's answer.
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.0
ok  that makes more sense now
 2 years ago

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

Jonask Group TitleBest ResponseYou've already chosen the best response.0
the easy element about this is that there are n constants
 2 years ago

experimentX Group TitleBest 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.
 2 years ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
no constants (coeffecient)\[\left[\begin{matrix}1& ...1\\ & \end{matrix}\right]=\]
 2 years ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
the coeffecient matrix
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
n = N+2, \[N + N(N1)/2\]
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
@experimentX for \(n=N+2\) that is not right.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
huh .. isn't it \[ \binom{N}{1} + \binom{N}{2}\]
 2 years ago

estudier Group TitleBest 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?
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
another un/lucky guess\[ \left(\begin{matrix}n1 \\ nN\end{matrix}\right) \]
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
@experimentX very close, but not right. Hope you will get it!
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
@estudier in your case N=2, n=3 N<n.
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
Ah, backwards ...duh!:)
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
@klimenkov what about n=N+2 case?
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
is that correct?
 2 years ago

asnaseer Group TitleBest 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?
 2 years ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
N=n >1 N=n1>1 N=n2>2 N=n3>3
 2 years ago

sauravshakya Group TitleBest 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
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
Wait a minute I think about your conclusions.
 2 years ago

asnaseer Group TitleBest 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
 2 years ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
That was only for n>5
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
Binomial coefficient, is it?
 2 years ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
@experimentX and @estudier I think question is related to the necklace question... right?
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.0
is it:\[1+F(Nn2)\]
 2 years ago

experimentX Group TitleBest 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. \[ \binom{N}{1} + \binom{N}{2} = N + {N(N1) \over 2} = {N(N+1) \over 2}\]
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.0
sorry, that should be:\[1+F(nN2)\]
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
@sauravshakya kind ov
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.0
that is for n>N+2
 2 years ago

klimenkov Group TitleBest 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.
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
Stars and Bars ex 0 > (N1 n1)
 2 years ago

asnaseer Group TitleBest 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. :)
 2 years ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
@klimenkov I know there must be a pattern but I dont think it is Fibonacci
 2 years ago

klimenkov Group TitleBest 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
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.0
These were my thoughts: dw:1350214566420:dw
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
My English is not too good. But I think @estudier should be given Best response.
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.0
Looking at this again, it looks more like:\[F(nN)+F(nN1)+...+F(1)\]
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.0
@estudier I did not understand what you meant by: Stars and Bars ex 0 > (N1 n1)
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
http://openstudy.com/users/klimenkov#/updates/50787c0ae4b02f109be43638 Try to get this.
 2 years ago

experimentX Group TitleBest 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
 2 years ago

estudier Group TitleBest 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.
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.0
Oh I see  thanks for clarifying.
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
The idea of linking binomial coefficient with Fibonacci is interesting....
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
shouldn't this be like this (N1 n1) > (n1 N1) ??
 2 years ago

klimenkov Group TitleBest 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?
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
that and this are equivalent \[ \left(\begin{matrix}n1 \\ nN\end{matrix}\right) \]
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
Can we use k and n, I am getting very confused:)
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
Shame on me. Somebody clear my Smartscore...
 2 years ago

experimentX Group TitleBest 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.
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
@experimentX was right since the very beginning.
 2 years ago

asnaseer Group TitleBest 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?
 2 years ago

estudier Group TitleBest 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)
 2 years ago

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

asnaseer Group TitleBest ResponseYou've already chosen the best response.0
D'oh! Now I think its time for someone to take my smartscore away! :)
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
Did anyone get this? Why does this formula give the right result?
 2 years ago

estudier Group TitleBest 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
 2 years ago

estudier Group TitleBest 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
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
thanks estudier for sharing!!
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.0
I also extends my thanks to estudier. :)
 2 years ago
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