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AravindG

find the no: of integral solutions of inequation x+y+z+u<=65

  • 2 years ago
  • 2 years ago

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  1. FoolForMath
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    0 is included?

    • 2 years ago
  2. FoolForMath
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    I can do this using binomial coefficients.

    • 2 years ago
  3. AravindG
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    i dont know tht fool

    • 2 years ago
  4. FoolForMath
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    Using stars and bars, the answer is given by \[ \huge \sum \limits_{n=0}^{65} \binom{n+3}{3} \]

    • 2 years ago
  5. satellite73
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    can't wait to see this

    • 2 years ago
  6. satellite73
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    stars and bars??

    • 2 years ago
  7. AravindG
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    hw?

    • 2 years ago
  8. FoolForMath
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    http://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)

    • 2 years ago
  9. moneybird
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    this question is like putting three bookmarks in a book

    • 2 years ago
  10. satellite73
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    how cool is that? stars and bars, nice visual!

    • 2 years ago
  11. FoolForMath
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    Yeah it's coool :D

    • 2 years ago
  12. AravindG
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    coool

    • 2 years ago
  13. shaan_iitk
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    One can also think logically and get the solution .. say you have x+y+z+u <= k (some integer) and you have to find the number of solutions then it is like finding the number of ways where k sweets can be distributed to 4 kids and total sweets distributed may not be equal to k. Now lets imagine a situation where we have 4 women and k men. How many ways of arranging them? (4+k)!/(4!*k!) Now say I say that all men who come to the left of any woman are given to that woman (not literally) .. then we have solved our distribution problem and hence our equation.. so I believe the answer should be 69!/(4!*65!)

    • 2 years ago
  14. FoolForMath
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    @shaan_iitk How do you think stars and bars work? That is exactly the same thing you elaborated.

    • 2 years ago
  15. shaan_iitk
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    means?

    • 2 years ago
  16. shaan_iitk
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    Sorry but I am not aware of stars and bars ..

    • 2 years ago
  17. FoolForMath
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    I have posted the url before.

    • 2 years ago
  18. shaan_iitk
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    okk .. I didn't look at that.. I only looked at your expression which was in summation form.. I thought logically we don't need that summation..

    • 2 years ago
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