Here's the question you clicked on:
Ishaan94
Number of ways of distributing \(n\) identical things among \(r\) persons when each person can get any number of things. @Zarkon
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 \(\binom{n+r-1}{r}\)
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.
http://en.wikipedia.org/wiki/Stars_and_bars_%28combinatorics%29
Thank you. Please medal each-other.