A community for students.
Here's the question you clicked on:
 0 viewing
klimenkov
 3 years ago
How many natural solutions does this equation have?
\[x_1+x_2+\ldots+x_N=n\]
klimenkov
 3 years ago
How many natural solutions does this equation have? \[x_1+x_2+\ldots+x_N=n\]

This Question is Closed

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

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

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

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

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

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

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

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

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

asnaseer
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1@asnaseer that was reply on Jonask's answer.

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

klimenkov
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0the easy element about this is that there are n constants

experimentX
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0no constants (coeffecient)\[\left[\begin{matrix}1& ...1\\ & \end{matrix}\right]=\]

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

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

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

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

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

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

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

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

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

sauravshakya
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1Wait a minute I think about your conclusions.

asnaseer
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0That was only for n>5

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

sauravshakya
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0sorry, that should be:\[1+F(nN2)\]

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

klimenkov
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0Stars and Bars ex 0 > (N1 n1)

asnaseer
 3 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
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0These were my thoughts: dw:1350214566420:dw

klimenkov
 3 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
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1http://openstudy.com/users/klimenkov#/updates/50787c0ae4b02f109be43638 Try to get this.

experimentX
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0Oh I see  thanks for clarifying.

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

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

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

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

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

experimentX
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1@experimentX was right since the very beginning.

asnaseer
 3 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
 3 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
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1Did anyone get this? Why does this formula give the right result?

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

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.0I also extends my thanks to estudier. :)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.