Let $L$ denote the event that his left neighbor has left and $R$
the event that his right neighbor has left. Then:
$$P\left(L\cap R\right)=P\left(L\mid R\right)P\left(R\right)=\frac{9}{23}\frac{10}{24}$$
The second factor because of $24$ cars $10$ leave and all cars have equal chances to be one them. The first factor likewise, but now of $23$ cars $9$ are selected to leave.
You have to account for pairs of adjacent empty spaces rather than consecutive empty spaces.
Let's focus on the four empty spaces that remain once the twelve cars have parked. We wish to find the probability that there are at least two adjacent empty spaces among those four.
A pair of adjacent empty spaces: There are $15$ places for a block of two empty spaces to begin. Once they have been selected, there are $\binom{14}{2}$ ways to select the positions of the other two empty spaces, giving an initial count of
$$\binom{15}{1}\binom{14}{2}$$
arrangements with four empty spaces that include two adjacent empty spaces.
However, we have counted those arrangements in which there are two pairs of adjacent empty spaces twice, once for each way of designating one of those pairs of adjacent empty spaces as the pair of adjacent empty spaces. We only want to count those arrangements once, so we must subtract those arrangements in which there are two pairs of adjacent empty spaces.
Two pairs of adjacent empty spaces: This can occur in two ways. The pairs can overlap, in which case there are three consecutive empty spaces, or be disjoint.
Two overlapping pairs of adjacent empty spaces: This includes a block of three consecutive adjacent empty spaces. The block must begin in one of the first $14$ positions. That leaves $13$ positions in which to place the remaining empty space. Thus, there are
$$\binom{14}{1}\binom{13}{1}$$
such arrangements.
Two disjoint pairs of adjacent empty spaces: We have $14$ objects to arrange, two blocks of two empty spaces and $12$ occupied spaces. Choose which two of those $14$ positions will be filled with the blocks, which can be done in
$$\binom{14}{2}$$
ways.
If we subtract those arrangements in which there are two pairs of adjacent empty spaces from the total, we will not have counted those arrangements in which there are three pairs of adjacent empty spaces at all. This is because we first added them three times, once for each way we could designate one of those three pairs as the pair of adjacent empty spaces, and subtracted them three times, once for each way of the $\binom{3}{2}$ ways we could designate two of those three pairs as the pairs of adjacent empty spaces. Thus, we must add those arrangements with three pairs of adjacent empty spaces to the total.
Three pairs of adjacent empty spaces: This can only occur if there are four consecutive empty spaces. A block of four consecutive empty spaces must begin in one of the first $13$ positions.
Since there are $\binom{16}{4}$ ways to select four empty parking places, the probability that Auntie Em can park is
$$\frac{\dbinom{15}{1}\dbinom{14}{2} - \dbinom{14}{1}\dbinom{13}{1} - \dbinom{14}{2} + \dbinom{13}{1}}{\dbinom{16}{4}}$$
Best Answer
Lets split into cases depending on which parking spot is open after the four cars have parked. Number the cars $1-5$ and suppose car $5$ is the one that didn't show up.
Case $1$: Parking spot $5$ is left empty.
In this case there are $D_4=9$ ways for the four cars to park in the other four spots.
Case $2$: Parking spot $1$ is left empty.
Here, if car $1$ parks in spot $5$, then the other three cars fill in with $D_3=2$ ways. If car $1$ parks in any of the other three spots, you can check that there are $3$ ways cars $2-4$ can fill in, making $9$ total. For this case, there are $2+9=11$ ways for the cars to park.
Since parking spots $2-4$ are the same as parking spot $1$ by symmetry, we get $11$ ways for each of those cases too. Altogether we find $9+4\cdot11=\boxed{53}$ ways for the cars to park.