A parking lot has $16$ spaces in a row. Twelve cars arrive, each of which requires one parking space, and their drivers chose spaces at random from among the available spaces. Auntie Em then arrives in her SUV, which requires $2$ adjacent spaces. What is the probability that she is able to park?
This problem can be modeled as counting number of integer solutions to an indeterminate equation, together with the counting-by-the-opposite technique.
Instead of investigating when Auntie Em can park her car, let's examine when she cannot. It is clear that such situation arises when all available parking spaces are separated by one or more vehicles.
The diagram above shows that four spaces are separated by some cars. Each variable denotes the number of cars in that block. Clearly there are totally $12$ cars. Among these variables, $x_1$ and $x_5$ can be zero. Such case means that an open space is at the start or end of this parking lot. Meanwhile, $x_2$, $x_3$, and $x_4$ must be positive because otherwise two vacant spaces can join each other.
Hence, this problem is equivalent to counting the integer solutions to $$x_1 + x_2 + x_3 + x_4 + x_5 =12$$
where $x_1$, $x_5 \ge 0$ and $x_2$, $x_3$, $x_4 > 0$. The answer to this problem is the same as # 4135, i.e. $C_{13}^{4}$.
In order to calculate the probability, we also need to find the number of all possibilities where $12$ cars are parked in a $16$-space parking lot. This is simply equal to $C_{16}^{12}$. Therefore, the probability of Auntie Em $\textbf{can}$ park her car equals $$1-\frac{C_{13}^{4}}{C_{16}^{12}}=\boxed{\frac{17}{28}}$$