anonymous
  • anonymous
Number of ways of distributing \(n\) identical things among \(r\) persons when each person can get any number of things. @Zarkon
Mathematics
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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 \(\binom{n+r-1}{r}\)
Zarkon
  • Zarkon
stars and bars ;)
KingGeorge
  • KingGeorge
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.

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Zarkon
  • Zarkon
http://en.wikipedia.org/wiki/Stars_and_bars_%28combinatorics%29
anonymous
  • anonymous
Thank you. Please medal each-other.

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