Method 1: We can arrange the eight girls in a row in $8!$ ways. This creates nine spaces in which to insert the boys, seven between successive girls and two at the ends of the row.
$$\square g \square g \square g \square g \square g \square g \square g \square g \square$$
To ensure that no two of the boys sit in consecutive seats, we choose six of these nine spaces in which to insert the boys, which we can do in $\binom{9}{6}$ ways. The boys can be arranged in the six selected spaces in $6!$ orders, so the number of seating arrangements of eight girls and six boys in which no two of the boys sit in consecutive seats is $$8!\binom{9}{6}6!$$ as you found.
Method 2: This is a modification of your approach.
We can arrange the six boys in a row in $6!$ ways. This creates seven spaces in which to place the girls, five between successive boys and two at the ends of the row.
$$\square b \square b \square b \square b \square b \square b \square$$
If $x_k$ represents the number of girls placed in the $k$th space, then
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 8 \tag{1}$$
The requirement that no two boys sit in adjacent seats means that $x_2, x_3, x_4, x_5, x_6 \geq 1$, while $x_1, x_7 \geq 0$. If we let $y_k = x_k - 1$, $2 \leq k \leq 6$, then $y_k$ is a nonnegative integer. Substituting $y_k + 1$ for $x_k$, $2 \leq k \leq 6$, in equation 1 yields
\begin{align*}
x_1 + y_1 + 1 + y_2 + 1 + y_3 + 1 + y_4 + 1 + y_5 + 1 + y_6 + 1 + x_7 & = 8\\
x_1 + y_1 + y_2 + y_3 + y_4 + y_5 + y_6 + x_7 & = 3 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers. A particular solution corresponds to the insertion of six addition signs in a row of three ones. The number of such solutions is
$$\binom{6 + 3}{6} = \binom{9}{6}$$
since we must choose which three of the nine positions (three ones and six addition signs) will be filled with additions signs. Finally, we can arrange the eight girls in the selected positions in $8!$ ways, again yielding $$6!\binom{9}{3}8!$$
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
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 $$\binom{10}{4} - \binom{9}{1}\binom{8}{2} + \binom{8}{1}\binom{7}{1} + \binom{8}{2} - \binom{7}{1}$$