What is wrong with my approach for CLRS 5.4-6 : Given n balls and n bins,find expected number of empty bins? I am trying to find expected number of empty bins after n balls are tossed into n bins. And each toss is  independent and equally likely to end up in any bin. Below is my approach.
My indicator variable is 
$X_i$ : i bins are empty
$$ Pr[X_i]=  \frac{\binom{n}{n-i} * n^\left(n-i\right)}{n^n}$$
And excepted number of empty bins is :
$$
\sum_{i=1}^\left(n-1\right) i*Pr[X_i]
$$
After simplifying the above equation I get:
$$
\sum_{i=1}^\left(n-1\right) \frac{\left(n-1\right)!}{\left(i-1\right)!*\left(n-i\right)!*n^\left(i-1\right)}
$$
But in the solution that I found on the web, the indicator variable is : Let Xi be the event that all the balls fall in bins, other than the ith. And then expected number of empty bins is :
$$
\sum_{i=1}^n \left(\frac{n-1}{n}\right)^n
$$
But according to me the the indicator variable chosen above is wrong. As they are adding the probabilities that ith bin is empty.So at a time only one bin is considered empty. Whereas there can be more than one bin empty at a time.
Is there something wrong with my understanding of the above problem?
 A: The web solution is correct; it uses linearity of expectation.  Suppose we label the boxes $1$ through $n$, and we determine $X_1$, the expected number of boxes numbered $1$ that are empty.  Since there is obviously only one box numbered $1$, this value is equal to the probability that that one box is empty:
$$
E(X_1) = P(\text{box $1$ is empty}) = \left(\frac{n-1}{n}\right)^n
$$
By symmetry, we have $E(X_1) = E(X_2) = \cdots = E(X_n)$, and then by linearity of expectation, the expected number of boxes that are empty is
\begin{align}
E(X) & = E(X_1+X_2+\cdots+X_k) \\
     & = E(X_1)+E(X_2)+\cdots+E(X_n) \\
     & = nE(X_1) \\
     & = \frac{(n-1)^n}{n^{n-1}}
\end{align}

You can also proceed the way you set out.  However, in the expression you give for the probability that exactly $k$ boxes are empty, the numerator $\binom{n}{k} n^{n-k}$ does not, unfortunately, count only those cases; it also includes cases where more than $k$ boxes are empty, too (and overcounts them, to boot).
The correct expression is not trivial; it is
$$
\frac{n!}{k!} S(n, n-k)
$$
where $S(\cdot, \cdot)$ are Stirling numbers of the second kind.  See this OEIS entry for more details on the particular count.  It is also written
$$
\frac{n!}{k!} \left\{ n \atop n-k \right\}
$$
All in all, you're better off using linearity of expectation. :-)
