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.
 A: We have $10$ cars. We run each car and test its mpg. Each car fall into a slot with a certain probability, and so occupies one digit of $4444555551$.
So after $10$ cars have been tested, we have a string such as $1555414554$.
The number of the strings of this form is given by:
$$\dfrac{10!}{6!3!1!}$$
and the probability for each digit to occur is:
$$(0.4)^3, (0.5)^6, (0.1)^1$$
A: 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
$\dfrac{n!}{x!y!z!}\cdot p^x\cdot q^y\cdot r^z$
A: You can count in three stages and then multiply the counts to get a total.
First stage: In how many ways could you choose $3$ cars with less than $22$ mpg out of a total of $10$? In $\displaystyle \binom{10}{3}$ different ways. After this there are $7$ cars left.
Second stage: In how many ways could you choose $6$ cars with mpg in $[22, 26]$ out of a total of $7$? In $\displaystyle \binom{7}{6}$ different ways. After this there are $1$ car left.
Third stage: In how many ways could you choose $1$ car with more than $26$ mpg out of a total of $1$? In $1$ way.
Now, the total count is
$$\binom{10}{3}\binom{7}{6}(1) = \binom{10}{3}\binom{7}{6},$$
which is equal to the fancier multinomial notation
$$\binom{10}{3,6,1} = \frac{10!}{3!6!1!}.$$
Here, the question "In how many ways could you choose $x$ cars with certain property $P$ out of total of $y$?" could be read as "In how many ways could occur that there are $x$ cars with certain property $P$ out of total of $y$?". The property $P$ makes reference to the mpg range.
A: The trivariate multinomial distribution looks rather like a Binomial Distribution.
$$\mathsf P(X{=}x, Y{=}y, Z{=}n{-}x{-}y) = \dfrac{n!\;{p_{\!\lower{0.5ex}X}}^x\,{p_{\!\lower{0.5ex}Y}}^y\,{p_{\!\lower{0.5ex}Z}}^{n-x-y}}{x!\, y!\,(n-x-y)!}$$
Where $X,Y,Z$ are the random variables, ${p_{\!\lower{0.5ex}X}}, {p_{\!\lower{0.5ex}Y}}, {p_{\!\lower{0.5ex}Z}}$ are their probabilities, $n$ is the number of trials, and $(x,y)$ are the realised values within the integer support: $\{0\leq x\leq n\}{\times}\{ 0\leq y\leq n-x\}$

What you are counting are ways to obtain the favoured counts of $x$ category 1, $y$ category 2, and $n-x-y$ category 3 cars out of all the possible ways to do so.   This follows the same principle as for the Binomial Distribution.
There is a probability ${p_{\!\lower{0.5ex}X}}^x\,{p_{\!\lower{0.5ex}Y}}^y\,{p_{\!\lower{0.5ex}Z}}^{n-x-y}$ of obtaining these results in a particular arrangement, and there are $\dfrac{n!}{x!\, y!\,(n-x-y)!}$ distinct permutations of that arrangement.
A: The probability of having $m$ of the $n$ cars in the first category is 
$${n \choose m} 0.4^m 0.6^{n-m}.$$
This is just the binomial distribution formula.
Conditioned on being in one of the second two categories, the probability of being in the second category is $\frac{0.5}{0.6}$ and the probability of being in the third category is $\frac{0.1}{0.6}$. So using the binomial distribution again, the probability of $k$ of the remaining $n-m$ cars being in the second category is 
$${n-m \choose k} \frac{0.5^k 0.1^{n-m-k}}{0.6^{n-m}}.$$
Because we've conditioned, you can now just multiply these and simplify, getting the "trinomial" distribution formula:
$${n \choose m} {n-m \choose k} 0.4^m 0.5^k 0.1^{n-m-k}.$$
You can prove the general multinomial formula by an inductive version of this argument. You can also simplify the products of the binomial coefficients: for instance here ${n \choose m} {n-m \choose k} = \frac{n!}{m! (n-m)!} \frac{(n-m)!}{k! (n-m-k!)}=\frac{n!}{m! k! (n-m-k)!}$. It is not hard to see algebraically that this cancellation will always occur. You can see this by a combinatoric argument as well: the number of ways to take $n$ things and put $m$ of them in one group, $k$ in another, and $n-m-k$ in a third is the number of ways to order all $n$ things divided by the number of ways to reorder the things within the groups.
