Multinomial probability confusion The probabilities are $0.4$, $0.5$, and $0.1$ that, in a city driving, a certain compact car will average less than $22$ miles per gallon, from $22$ to $26$ miles per gallon, or more than $26$ miles per gallon. Find the probability that among $10$ such cars tested, $3$ will average $22$ miles per gallon, $6$ will average $22$ to $26$ miles per gallon, and $1$ will average more than $26$ miles per gallon.
I know we should multiply the probabilities, $(0.5)^{6}$ and such, to the cars we are choosing, but how exactly to choose the cars I am confused. 
EDIT: I do not want the formula as I already know it exists. I want an argument that counts what the question is asking for. 
Can anyone give me a hint on how to get started? I'm writing things down on paper, but it doesn't get me anywhere.

It is just an extension of the binomial distribution.
In $n$ trials, an event occurs $x$ times with probability $p$, $y$ times with probability $q$,
with $x+y=n, p+q=1$, 
and we could as well write the binomial coefficient $\dbinom{n}{x}$ as a permutation, $\dfrac{n!}{x!y!}$, yielding
$\dfrac{n!}{x!y!}\cdot p^x\cdot q^y$
Extend it to, say, $3$ possibilities for each event $x,y,z$ with probabilities $p,q,r$ respectively, 
$x+y+z=n, p+q+r=1$, and we analogously get
$\dfrac{n!}{x!y!z!}\cdot p^x\cdot q^y\cdot r^z$
