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

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

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

klimenkov
 2 years ago
Best 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.

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

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

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

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

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

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

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

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

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

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

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

asnaseer
 2 years ago
Best 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?

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

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

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

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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.3for n=N there should be 1 solutions. for n=N+1 there should be N solutions.

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

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

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

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

estudier
 2 years ago
Best 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?

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

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

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

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

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

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

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

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

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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.0I 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

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

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

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

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

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

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

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

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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.0Wait  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. :)

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

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

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

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

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

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

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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.3for 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

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

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

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

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

klimenkov
 2 years ago
Best 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?

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

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

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

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

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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.0I 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?

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

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

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

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

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

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

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

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