What is the probability that each of the vehicles will be made to carry at least one local tourist? Three vehicles (one blue, one green and one grey) with a carrying capacity of 8 passengers each are to be used to ferry 18 international tourists and 5 local tourists (who are a family) from OR Tambo Airport to Vilakazi Street, Soweto. 
If the logistics manager randomly assigns the tourists to the vehicles and makes a decision that the blue vehicle will carry 7 tourists, what is the probability that each of the vehicles will be made to carry at least one local tourist?
Answer: $\frac{2194901280}{3155170590} = \frac{16}{23} = 0.69565$
I can get the denominator: ${23\choose7}{16\choose8}{8\choose8} = 3155170590$
The numerator/combinatorics part is the one that I'm struggling with.
How do you assign at least one local tourist to each vehicle and assign the rest of the people to those same vehicles at the same time?
 A: The numerator can be calculated by the counting the number of ways not to get at least one local per bus. That is, by counting the number of ways to get at least one bus without a local. We use the inclusion-exclusion priniple.
Let $N_b$ be the number of ways the blue bus has no locals, $N_g$ for the green bus, $N_y$ for the grey bus.
Let $N_{bg}$ be the number of ways both the blue and green buses have no locals, and $N_{by}$ and $N_{gy}$ similarly defined.
Let $N$ be the total number of ways at least one bus has no locals.
$N_b = \binom{18}{7} \binom{16}{8}$ (choosing $7$ non-locals from a possible $18$ then a further $8$  of any of the remaining $16$ passengers).
Similarly, $N_g = N_y = \binom{18}{8} \binom{15}{8}$.
$N_{bg} = N_{by} = \binom{18}{7} \binom{11}{8}$ (choosing $7$ non-locals from a possible $18$ then a further $8$ non-locals from the remaining $11$ of them).
Similarly, $N_{gy} = \binom{18}{8} \binom{10}{8}$.
$N_{bgy} = 0$ because at least one bus must have locals.
The I-E formula:
\begin{eqnarray*}
N &=& \left(N_b+N_g+N_y\right) - \left(N_{bg}+N_{by}+N_{gy}\right) + \left(N_{bgy}\right) \\
&& \\
&=& \left[\binom{18}{7} \binom{16}{8} + 2\binom{18}{8} \binom{15}{8}\right] - \left[2\binom{18}{7} \binom{11}{8} + \binom{18}{8} \binom{10}{8}\right] + 0 \\
&& \\
&=& 960269310
\end{eqnarray*}
We want the opposite of that: the number of ways all buses have at least one local so we subtract this value from the total number of ways to arrange the passengers. This number is the denominator, which you already calculated.
So the numerator $= 3155170590 - 960269310 = 2194901280$.
