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dan815

  • one year ago

a+b+c+d=k a,b,c,d>=1 can you find a formula for the number of integer solutions as a function of k?

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  1. anonymous
    • one year ago
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    It looks like repeated choose operation.

  2. ganeshie8
    • one year ago
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    \[\binom{k-1}{3}\]

  3. dan815
    • one year ago
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    how did u get that so fast lol

  4. ganeshie8
    • one year ago
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    it is a good ol stars and bars problem \(k\) stars and \(3\) bars

  5. dan815
    • one year ago
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    wait wont u have a 0 in the middle with that form

  6. dan815
    • one year ago
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    oh did u subtract 3 spots

  7. dan815
    • one year ago
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    for the 3 zeros not allowed

  8. ganeshie8
    • one year ago
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    \[\star\star|\star\star|\star|\star\star\]

  9. dan815
    • one year ago
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    like if zeros are allowed it would be K+3 choose 3 right

  10. ganeshie8
    • one year ago
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    Yep

  11. dan815
    • one year ago
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    how come this works, so u subtract 4 spots? for the 4 zeros no allowed

  12. ganeshie8
    • one year ago
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    Firstly, notice that those 3 bars divide the 7 stars into 4 pieces

  13. dan815
    • one year ago
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    right

  14. ganeshie8
    • one year ago
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    You want to have at least 1 star in each piece, yes ?

  15. dan815
    • one year ago
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    right

  16. dan815
    • one year ago
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    okay i see now that limits the total number of places

  17. dan815
    • one year ago
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    ahhh

  18. ganeshie8
    • one year ago
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    thats it! so only the places between stars are allowed

  19. dan815
    • one year ago
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    dang that was simple lol

  20. dan815
    • one year ago
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    i got to this point with some other little trick lol

  21. dan815
    • one year ago
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    like u know if u look at the difference of the solutions that has a stars and bar solution again allowing 0

  22. ganeshie8
    • one year ago
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    solving a+b+c+d = k over nonnegative integers ?

  23. dan815
    • one year ago
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    kinds cool that exists

  24. dan815
    • one year ago
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    yeah like umm

  25. dan815
    • one year ago
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    say we were looking at a+b+c=7 then the simple case 1+1+5=7 now i looked at their differences

  26. dan815
    • one year ago
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    |dw:1435697202677:dw|

  27. dan815
    • one year ago
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    you can do stars and bars on the differences to see all the possible differences

  28. dan815
    • one year ago
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    only thing is you have to add 1 extra spot in for the differences

  29. ganeshie8
    • one year ago
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    how do you know there exists a bijection between both the worlds

  30. dan815
    • one year ago
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    well the way i see it is if u write the 1 ,1 ,5 as 1,1,1 +4 now that +4 can be distributed in any way about the 3 terms

  31. dan815
    • one year ago
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    so a+b+c=k can we rewritten as 1+1+(1+k-3) and the k-3 can be distributed in any way allowing 0 about the 3 terms

  32. ganeshie8
    • one year ago
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    Ahh I see so that brings in the evil paritions

  33. dan815
    • one year ago
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    yaa

  34. dan815
    • one year ago
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    ill post the modified peuler question, i think there are still some cool tricks to be discovered with how to reduce the cases efficiently

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