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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

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  1. Jonask Group Title
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    \[N\]

    • 2 years ago
  2. experimentX Group Title
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    is N<n ??

    • 2 years ago
  3. klimenkov Group Title
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    Very interesting answer, because doesn't depend on \(n\).

    • 2 years ago
  4. Jonask Group Title
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    is that a linear equation,\[x+y+z=b\] how many solutions

    • 2 years ago
  5. klimenkov Group Title
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    @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
  6. experimentX Group Title
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    |dw:1350212084534:dw|

    • 2 years ago
  7. asnaseer Group Title
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    if \(x_i\) represents ANY real number, then there are an infinite number of solutions

    • 2 years ago
  8. experimentX Group Title
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    |dw:1350212197575:dw|

    • 2 years ago
  9. Jonask Group Title
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    the problem is the difference between the solution of the equation and the solution set

    • 2 years ago
  10. klimenkov Group Title
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    @asnaseer I wrote in the statement of the problem that xi are natural.

    • 2 years ago
  11. experimentX Group Title
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    |dw:1350212336061:dw|

    • 2 years ago
  12. asnaseer Group Title
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    ah! ok - thanks for pointing that out.

    • 2 years ago
  13. Jonask Group Title
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    |dw:1350180042769:dw|slution set implies tha t there are infinite solutions to the numbers \[x _{1}+x _{2}=2\]

    • 2 years ago
  14. Jonask Group Title
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    (0,2) (2,0)\ (1,1)

    • 2 years ago
  15. Jonask Group Title
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    i only see 3 positive ones

    • 2 years ago
  16. klimenkov Group Title
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    @Jonask zero is not a natural number.

    • 2 years ago
  17. experimentX Group Title
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    is it ?? \[ \binom {n+N-1}{n-N}\]

    • 2 years ago
  18. asnaseer Group Title
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    @klimenkov are you saying that you know what the answer is and that it does NOT depend on n?

    • 2 years ago
  19. klimenkov Group Title
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    @asnaseer that was reply on Jonask's answer.

    • 2 years ago
  20. asnaseer Group Title
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    ok - that makes more sense now

    • 2 years ago
  21. klimenkov Group Title
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    @experimentX use your method to find the number of solutions for \(x_1=2\).

    • 2 years ago
  22. Jonask Group Title
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    the easy element about this is that there are n constants

    • 2 years ago
  23. experimentX Group Title
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    for n=N there should be 1 solutions. for n=N+1 there should be N solutions.

    • 2 years ago
  24. Jonask Group Title
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    no constants (coeffecient)\[\left[\begin{matrix}1& ...1\\ & \end{matrix}\right]=\]

    • 2 years ago
  25. Jonask Group Title
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    the coeffecient matrix

    • 2 years ago
  26. experimentX Group Title
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    n = N+2, \[N + N(N-1)/2\]

    • 2 years ago
  27. klimenkov Group Title
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    @experimentX for \(n=N+2\) that is not right.

    • 2 years ago
  28. experimentX Group Title
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    huh .. isn't it \[ \binom{N}{1} + \binom{N}{2}\]

    • 2 years ago
  29. estudier Group Title
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    "there are no solutions if N>n" x_1 + x_2 = 3 (answer 1 +2) Am I missing something?

    • 2 years ago
  30. experimentX Group Title
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    another un/lucky guess\[ \left(\begin{matrix}n-1 \\ n-N\end{matrix}\right) \]

    • 2 years ago
  31. klimenkov Group Title
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    @experimentX very close, but not right. Hope you will get it!

    • 2 years ago
  32. klimenkov Group Title
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    @estudier in your case N=2, n=3 N<n.

    • 2 years ago
  33. estudier Group Title
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    Ah, backwards ...duh!:-)

    • 2 years ago
  34. experimentX Group Title
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    @klimenkov what about n=N+2 case?

    • 2 years ago
  35. experimentX Group Title
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    is that correct?

    • 2 years ago
  36. asnaseer Group Title
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    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
  37. sauravshakya Group Title
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    N=n --->1 N=n-1-->1 N=n-2--->2 N=n-3--->3

    • 2 years ago
  38. sauravshakya Group Title
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    N=1 --->1 N=n/2 if n is even and (n-1)/2 if n is odd

    • 2 years ago
  39. klimenkov Group Title
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    Wait a minute I think about your conclusions.

    • 2 years ago
  40. asnaseer Group Title
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    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
  41. sauravshakya Group Title
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    That was only for n>5

    • 2 years ago
  42. estudier Group Title
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    Binomial coefficient, is it?

    • 2 years ago
  43. sauravshakya Group Title
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    @experimentX and @estudier I think question is related to the necklace question... right?

    • 2 years ago
  44. asnaseer Group Title
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    is it:\[1+F(N-n-2)\]

    • 2 years ago
  45. estudier Group Title
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    N-1,n-1

    • 2 years ago
  46. experimentX Group Title
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    \[ \binom{n-1}{N-1} = \left(\begin{matrix}N+2-1 \\ N-1\end{matrix}\right) = \binom{N+1}{N-1} = {(N+1)N \over 2} \] both seems to be equal here on this particular case. \[ \binom{N}{1} + \binom{N}{2} = N + {N(N-1) \over 2} = {N(N+1) \over 2}\]

    • 2 years ago
  47. asnaseer Group Title
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    sorry, that should be:\[1+F(n-N-2)\]

    • 2 years ago
  48. experimentX Group Title
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    @sauravshakya kind ov

    • 2 years ago
  49. asnaseer Group Title
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    that is for n>N+2

    • 2 years ago
  50. klimenkov Group Title
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    Sorry guys, the are so many answers so I can't quickly answer. I must think about them. Especially about Fibonacci.

    • 2 years ago
  51. estudier Group Title
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    Stars and Bars ex 0 -> (N-1 n-1)

    • 2 years ago
  52. asnaseer Group Title
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    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. :)

    • 2 years ago
  53. sauravshakya Group Title
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    @klimenkov I know there must be a pattern but I dont think it is Fibonacci

    • 2 years ago
  54. klimenkov Group Title
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    I know the answer and it doesn't equal to Fibonacci number +1. Write the way you think @asnaseer

    • 2 years ago
  55. asnaseer Group Title
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    These were my thoughts: |dw:1350214566420:dw|

    • 2 years ago
  56. klimenkov Group Title
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    My English is not too good. But I think @estudier should be given Best response.

    • 2 years ago
  57. asnaseer Group Title
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    Looking at this again, it looks more like:\[F(n-N)+F(n-N-1)+...+F(1)\]

    • 2 years ago
  58. asnaseer Group Title
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    @estudier I did not understand what you meant by: Stars and Bars ex 0 -> (N-1 n-1)

    • 2 years ago
  59. klimenkov Group Title
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    http://openstudy.com/users/klimenkov#/updates/50787c0ae4b02f109be43638 Try to get this.

    • 2 years ago
  60. experimentX Group Title
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    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

    • 2 years ago
  61. estudier Group Title
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    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
  62. asnaseer Group Title
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    Oh I see - thanks for clarifying.

    • 2 years ago
  63. estudier Group Title
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    The idea of linking binomial coefficient with Fibonacci is interesting....

    • 2 years ago
  64. experimentX Group Title
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    shouldn't this be like this (N-1 n-1) -> (n-1 N-1) ??

    • 2 years ago
  65. klimenkov Group Title
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    @experimentX yes. The answer is \(\left(\begin{matrix}n-1 \\ N-1\end{matrix}\right)\). Who will prove it?

    • 2 years ago
  66. experimentX Group Title
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    that and this are equivalent \[ \left(\begin{matrix}n-1 \\ n-N\end{matrix}\right) \]

    • 2 years ago
  67. estudier Group Title
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    Can we use k and n, I am getting very confused:-)

    • 2 years ago
  68. klimenkov Group Title
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    Shame on me. Somebody clear my Smartscore...

    • 2 years ago
  69. asnaseer Group Title
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    :)

    • 2 years ago
  70. experimentX Group Title
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    we are just putting n=n-N in your previous Q's answer ... whatever there is ... it is in your previous answer.

    • 2 years ago
  71. klimenkov Group Title
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    @experimentX was right since the very beginning.

    • 2 years ago
  72. asnaseer Group Title
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    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?

    • 2 years ago
  73. estudier Group Title
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    For positive n, k the number of different n-tuples (ex 0) with sum k is (k-1,n-1)

    • 2 years ago
  74. klimenkov Group Title
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    @asnaseer \[\left(\begin{matrix}5-1 \\ N-1\end{matrix}\right)=\frac{ 4! }{ (5-N)!(N-1)! }\]

    • 2 years ago
  75. asnaseer Group Title
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    D'oh! Now I think its time for someone to take my smartscore away! :)

    • 2 years ago
  76. klimenkov Group Title
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    Did anyone get this? Why does this formula give the right result?

    • 2 years ago
  77. estudier Group Title
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    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
  78. estudier Group Title
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    There is a proof on page 12 of http://www.ma.utexas.edu/users/geir/teaching/m390c/lecturenotes.pdf

    • 2 years ago
  79. experimentX Group Title
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    thanks estudier for sharing!!

    • 2 years ago
  80. estudier Group Title
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    yw:-)

    • 2 years ago
  81. asnaseer Group Title
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    I also extends my thanks to estudier. :)

    • 2 years ago
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