7 distinct trucks are sent to 3 different cities A,B,C.what is the number of possibilities? If exactly 2 trucks were sent to city A ,and exactly 4 trucks for city B and exactly 1 truck for city C
here is my thought process ,i looked at city C first and said there is 7 different possibilities ,then i looked at A and said to my self there is C(6,2) different possibilities ,and then only C(4,4) for B.
so the final count should be 7*C(6,2)*C(4,4) ,is this correct?
if not can you advise me on how to approach these kind of problems?;
i'm new to combinatorics.
 A: Your thought process is indeed correct. Expressing your solution in a slightly different fashion we could also say:
There $7$ trucks to choose from and we want to choose $2$ for $A$, out of the remaining $5$ trucks we want to choose $4$ for $B$ and we are left with $1$ truck that must go to $C$.
Then expressing our choices as binomial coefficients we have
$${7 \choose 2} \cdot {5 \choose 4} \cdot {1 \choose 1}.$$
We can see both of our results are the same when multiplied out (yielding $105$) so it does not matter which city we begin with concerning the selection of our $k$ trucks (where $k$ is the number of objects chosen in $n \choose k$).
Note: We are multiplying the result of each selection by the other selections in accordance to the rule of product, which states that if there $x$ ways of performing an action and $y$ ways of performing another action, there are $x\cdot y$ ways of performing both actions.
A: It is correct, however your notation is different from the standard mathematical notation.
The normally used notation for $c(a,b)$ is  $\binom {a}{b}$ ($b$ objects distributed over $a$ positions.)
The result of the operation is computed thus:
$b! / a! / (b-a)!$ where each number $n!$ (followed by an exlamation mark)  $= 1*2*3*... (n-3) * (n-2) * (n-1) * n$. It can also be expressed in words thusly: "Multiply the $1:th$ to $b:th$ highest integers in the range $1$ to $a$. Divide by integers $1$ to $b$."
In your example, the formula to compute the answer can be expanded as:
$$\frac{(7*6*5*4)*(3*2)*(1)}{(4*3*2*1)*(2*1)*(1)}$$
Which I hope will make it obvious that (as mentioned in the reply above) the order in which you multiply the results doesn't matter, because the order in which you divide the numbers within parenthesis above doesn't matter. If you changed the order:
$$\frac{(7*6)*(5*4*3*2)*(1)}{(2*1)*(4*3*2*1)*(1)}$$
...The result would still be the same.
