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mathmath333

  • one year ago

Counting problem

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{Find the number of natural number solutions of.}\hspace{.33em}\\~\\ & a+2b+c=100 \hspace{.33em}\\~\\ \end{align}}\)

  2. dan815
    • one year ago
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    hmm okay

  3. dan815
    • one year ago
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    b can be all even numbers

  4. dan815
    • one year ago
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    do you know how to solve the simpler version of this problem a+b+c=100

  5. mathmath333
    • one year ago
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    do i need to change 2b to d

  6. dan815
    • one year ago
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    sry i mean 2b is all even numbers

  7. mathmath333
    • one year ago
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    99C2

  8. dan815
    • one year ago
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    yeah thats right

  9. mathmath333
    • one year ago
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    i meant to use substitution as 2b=d

  10. dan815
    • one year ago
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    here is another way u can think about it a+b+b+c = 100 how would you solve this a+b+c+d=100

  11. mathmath333
    • one year ago
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    99C3

  12. dan815
    • one year ago
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    i think you would solve a+b+c=100 and divide by 2

  13. dan815
    • one year ago
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    as all the odd possibilities would not possible

  14. mathmath333
    • one year ago
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    u mean 99C2/2

  15. dan815
    • one year ago
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    yeah but let me think that might not technically be true we dont know if odd and even solutions that add to 100 is exactly equal or not

  16. mathmath333
    • one year ago
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    How about calculatinb a+b1+b2+d=100 and substracting something

  17. mathmath333
    • one year ago
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    ?

  18. mathmath333
    • one year ago
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    what do u mean

  19. dan815
    • one year ago
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    ok back

  20. dan815
    • one year ago
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    okay well maybe we can go back to the stars and bars method after this but i think we can do it with just sums

  21. dan815
    • one year ago
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    no wait there are 101 WAYs actually to add upto 100 when u pick 2b=0

  22. dan815
    • one year ago
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    |dw:1440274661135:dw|

  23. dan815
    • one year ago
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    does 0 count as natural number though?

  24. mathmath333
    • one year ago
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    No

  25. Zarkon
    • one year ago
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    sometimes...some define it to include 0 and others don't

  26. anonymous
    • one year ago
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    You could brute force it, write a python program to count it .

  27. dan815
    • one year ago
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    okay then lets subtract that out

  28. mathmath333
    • one year ago
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    0 is whole number

  29. mathmath333
    • one year ago
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    No pen paper method i need

  30. Zarkon
    • one year ago
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    http://mathworld.wolfram.com/NaturalNumber.html

  31. dan815
    • one year ago
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    |dw:1440274943096:dw|

  32. dan815
    • one year ago
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    2401

  33. dan815
    • one year ago
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    do you want to check with brute force

  34. mathmath333
    • one year ago
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    yea how do u brute force lol

  35. dan815
    • one year ago
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    tbh its just writing what i did up there in a code lol

  36. dan815
    • one year ago
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    i didnt really do it a very smart way summation is pretty bruteforcy

  37. dan815
    • one year ago
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    the answer is definately close to (99 choose 2)/2

  38. dan815
    • one year ago
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    because the odd + even combinations are not equal in this case there is going to be some fluctuation around that number

  39. mathmath333
    • one year ago
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    a short research shows a+2b+c=10--->4! ways a+2b+c=12--->5! ways a+2b+c=14--->6! ways Can i somehow get a pattern

  40. dan815
    • one year ago
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    @kainui can you see a way to do this problem combinatorically

  41. anonymous
    • one year ago
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    this would be the 'brute force' method Python code: count=0 for a in range(1,100): for b in range(1,100): for c in range(1,100): if a + 2*b+c ==100: count = count + 1 print count >>> 2401

  42. mathmath333
    • one year ago
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    yea 2401 is correct

  43. anonymous
    • one year ago
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    a+2b+c=10--->4! ways a+2b+c=12--->5! ways a+2b+c=14--->6! ways you can derive a pattern from this

  44. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{When a and c are even then}\hspace{.33em}\\~\\ & 2a'+2b+2c'=100\hspace{.33em}\\~\\ & \implies a'+b+c'=50\hspace{.33em}\\~\\ & \implies \dbinom{49}{2}\hspace{.33em}\\~\\ & \normalsize \text{When a and c are odd then}\hspace{.33em}\\~\\ & 2a'-1+2b+2c'-1=100\hspace{.33em}\\~\\ & \implies a'+b+c'=51\hspace{.33em}\\~\\ & \implies \dbinom{50}{2}\hspace{.33em}\\~\\ & \normalsize \text{Total}=\dbinom{49}{2}+\dbinom{50}{2}=2401\hspace{.33em}\\~\\ \end{align}}\)

  45. mathmath333
    • one year ago
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    that pattern was incorrect

  46. anonymous
    • one year ago
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    edit* actually the pattern should be a+2b+c=10--->4^2 ways a+2b+c=12--->5^2 ways a+2b+c=14--->6^2 ways you can derive a pattern from this a+2b+c=n--->(1/2*n -1)^2 ways

  47. anonymous
    • one year ago
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    when n = 100, 1/2*100 -1 = 49 and 49^2 = 2401

  48. mathmath333
    • one year ago
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    yea correct 49^2

  49. anonymous
    • one year ago
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    thats a nice solution you have so how do you know that a' + b + c ' = 50 has (49 choose 2 ) ways to find positive solutions

  50. mathmath333
    • one year ago
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    u have to use stars and bars method https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)

  51. anonymous
    • one year ago
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    interesting :)

  52. anonymous
    • one year ago
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    For any pair of positive integers n and k, the number of k-tuples of positive integers whose sum is n is equal to the number of (k − 1)-element subsets of a set with n − 1 elements. we had 50 is the sum , and 3 tuples so that means 49 choose 2

  53. anonymous
    • one year ago
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    And then consider the case when a,b are both even or both odd. Note that a,c cannot be one even and one odd, because odd + even + even = odd

  54. mathmath333
    • one year ago
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    yep

  55. mathmath333
    • one year ago
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    the language given is wikipdia is technical , if u have any doubt u can go through stars and bars videos on youtubes

  56. anonymous
    • one year ago
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    oh, do you have a link? or should i google

  57. mathmath333
    • one year ago
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    here is one of the links https://brilliant.org/discussions/thread/stars-and-bars/ or simply try google or youtube

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