I am trying to understand the accepted answer to this question: Find: The expected number of urns that are empty
And am stuck on the part I mentioned above. I understand that:
$\sum\limits_{i=1}^n \frac{i-1}{n}=\frac{0}{n}+\frac{1}{n}+...+\frac{n-1}{n}.$ But why does this equal $\frac{n\choose2}{n}$?
I have also seen this sum solved as follows:
$\sum\limits_{i=1}^n \frac{i-1}{n}=\frac{1}{n}\sum\limits_{i=1}^n i-1=\frac{1}{n}\sum\limits_{j=0}^{n-1}j=\frac{1}{n}\frac{n(n-1)}{2}=\frac{n-1}{2}$
With this method, I am having trouble seeing what exactly $j$ is and why we all of sudden entered it into the equation.
I'd appreciate if someone could shed light on either of these ways of computing the sum.
Thanks