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.

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.
