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.
It looks like you applied the Inclusion-Exclusion Principle to cases with two consecutive, three consecutive, and four consecutive numbers. However, you should instead apply the Inclusion-Exclusion Principle to pairs of consecutive numbers.
There are $\binom{10}{4}$ ways to choose four numbers from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.
A pair of consecutive numbers: Your count is correct. The smaller of the two consecutive numbers must occur in one of the first nine positions. Choosing the smaller also determines the larger. The remaining two numbers can be selected in
$\binom{8}{2}$ ways, so there are $$\binom{9}{1}\binom{8}{2}$$ such selections.
Two pairs of consecutive numbers: This can occur in two ways. The pairs can overlap, or they are disjoint.
Two overlapping pairs: This means that three consecutive numbers are selected. Since the smallest of these three consecutive numbers must occur in one of the first eight positions. That leaves seven choices for the remaining number. Hence, there are $$\binom{8}{1}\binom{7}{1}$$ such selections.
Two disjoint pairs: We have eight available positions, two for the pairs and six for the other six numbers. Choose two of the eight positions for the pairs. Doing so determines the pairs. For instance, if we choose the third and fifth positions, then the pairs are $3, 4$ and $6, 7$.
$$1, 2, \boxed{3, 4}, 5, \boxed{6, 7}, 8, 9, 10$$
Hence, there are $$\binom{8}{2}$$ such selections.
Three pairs: Since we are only selecting four numbers, this can only occur if we have four consecutive numbers. The smallest of these numbers can be selected in seven ways. Hence, there are $$\binom{7}{1}$$ such selections.
By the Inclusion-Exclusion Principle, the number of ways four numbers can be selected from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ such that no two consecutive numbers are selected is
$$\binom{10}{4} - \binom{9}{1}\binom{8}{2} + \binom{8}{1}\binom{7}{1} + \binom{8}{2} - \binom{7}{1}$$
Best Answer
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 $$\frac{\dbinom{15}{1}\dbinom{14}{2} - \dbinom{14}{1}\dbinom{13}{1} - \dbinom{14}{2} + \dbinom{13}{1}}{\dbinom{16}{4}}$$