Say there are $5$ distinguishable bins and one throws $7$ distinguishable balls. What is probability of exactly $2$ empty bins? The two methods below do not agree.
Method 1: inclusion exclusion \begin{align*} P_1 & = \frac{5C2 \cdot \{\text{number of ways to deposit $7$ balls in $3$ bins without any empty}\}}{5^7}\\ & = \frac{5C2 \cdot [3^7 - (3C1•2^7 - 3C2 • 1^7)]}{5^7}\\ & = \frac{18060}{5^7}\\ & = 0.231 \end{align*} The $5C2$ is the number of ways to choose the $2$ empty bins from $5$.
Method 2: Assume the probability a ball misses any $2$ bins of the $5$ total is $3/5$. Thus, the probability all $7$ balls miss these same $2$ bins is
$P_2 = (3/5)^7 = 0.0280$
Since same $5C2$ ways to choose any $2$ bins, the final probability is
$P_2 = 0.280$
Why do probabilities not agree ??
P2 over counts the cases where more than two bins are empty. If three bins are empty you will count it three times, once for each pair of empty bins. If four bins are empty you will count it six times. P1 carefully counts the cases where exactly two bins are empty.