Intuition for using the multinomial coefficient 
We have 9 parking spots, and 9 cars which will be parked randomly in these spots. 3 of the cars are sport cars, 3 of them trucks and 3 of them small cars. What is the probability that the 3 sport cars are parked next to each other?

So I already solved this using permutations once, and then again using combinations, but now I want to solve it using the multinomial coefficient. Let's call the event that the 3 sport cars are parked next to each other $A$. Then $$P(A) = \dfrac{n_A}{N}$$
$N=  \dbinom{9}{3,3,3}$, as we're using multinomial coefficients. 
Then $n_A = 7 \times \dbinom{6}{3,3}$ apparently, but I don't understand the $\dbinom{6}{3,3}$ term. Why does it matter how the remaining (non-sport) cars are ordered? I thought the point of combinations (binomial or multinomial) was that the order doesn't matter? 
 A: Let we number the parking slots with $1,2,\ldots,8,9$. If the three sport cars are next to each other, they occupy the slots $(1,2,3)$, $(2,3,4),\ldots,(7,8,9)$ (so $7$ possibilities) and they can be ordered in $3!=6$ ways. There are $6$ remaining slots: we may choose in $\binom{6}{3}=20$ ways the three slots occupied by trucks, then we have $3!$ orderings for the trucks and $3!$ orderings for the small cars. 
Let we count, now, the total number of arrangements (unrestricted). We have $\binom{9}{3}=84$ ways for choosing the positions of the sport cars, then $\binom{6}{3}=20$ ways for choosing the positions of the small cars, and $3!\cdot 3!\cdot 3!$ ways for ordering trucks, small cars and sport cars in their corresponding parking slots. It follows that the wanted probability is given by
$$ \frac{7\cdot 20\cdot 3!\cdot 3!\cdot 3!}{84\cdot 20\cdot 3!\cdot 3!\cdot 3!}=\frac{7}{84}=\frac{7}{\binom{9}{3}}=\color{red}{\frac{1}{12}}\approx 8.33\%. $$
A: "As above, so bellow."   When comparing counts of atoms in the favoured space (the numerator) and the total space (denominator), they must be counted in the same general manner. †   Don't compare apples and orang-utans.
[ † PS: All atomic outcomes must have equal probability for this to be a valid method for measuring probability of the event. ]
By no coincidence, these three methods measure the same probability:

There are $9!$ ways to arrange nine singletons in list.    There are $7!3!$ ways to arrange six singletons and one triple in a list when the triple must remain connected.
$$\dfrac{7!~3!}{9!}$$

There are $\binom 9 3$ ways to pick spots for the sports cars.   There are $7$ ways where these cars remain adjacent.
$$\dfrac{~7~}{{9!}/{6!~3!}}$$

There are $\binom{9}{3,3,3}$ ways to pick places for all the vehicles.   There are $7\binom{6}{3,3}$ ways to pick places for them such that the sports cars remain adjacent.
$$\dfrac{7\cdot{6!}/{3!~3!}}{{9!}/{3!~3!~3!}}$$
