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Ishaan94

  • 2 years ago

Number of ways of distributing \(n\) identical things among \(r\) persons when each person can get any number of things. @Zarkon

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  1. Ishaan94
    • 2 years ago
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    Usually it should be \(\binom{n}r\) but since each person can get any number of things including zero. I made it \(\binom{n+r}{r}\) but in the book it's given \(\binom{n+r-1}{r}\)

  2. Zarkon
    • 2 years ago
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    stars and bars ;)

  3. KingGeorge
    • 2 years ago
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    I believe this could be solved with the equation \[x_1+x_2+...+x_r=n.\]Where every \(x_i\) gets an integer greater than or equal to 0. Solving this problem is well known as the "Stars and Bars" problem and is given by \[\left(\!\!\binom{n}{r}\!\!\right)=\binom{n+r-1}{r}\] You have that \(-1\) because that's how many symbols you have if you write out every integer as a star, and every plus sign as a bar.

  4. Zarkon
    • 2 years ago
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    http://en.wikipedia.org/wiki/Stars_and_bars_%28combinatorics%29

  5. Ishaan94
    • 2 years ago
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    Thank you. Please medal each-other.

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