Understanding counting. I encountered this question recently:

Suppose there are 3 benches in the front row and 7 benches in the
  second row, how many ways a group of 10 children can be seated in such
  arrangement?

There were given two solutions to the same problem:


*

*Its an Arrangement hence there are $10!$ ways. The solution seem easy to understand.

*The other solution was like:


We can choose $3$ children for the first row in $10\choose 3$ ways, those $3$ can be arranged in $3!$ ways and the remaining $7$ can be arranged in $7!$ ways in the back row. So, the total number of ways are ${10 \choose 3}*3!*7!$ which equals $10!$.
I seriously got bowled over by the second solution, it went over my head like anything.
I am new to discrete math and as far as I know counting boils down to two simple rules sum and product. Both of the above solutions are using the same product rule but why I am failing to understand the second one?
Question
I know the product rule but still the multiplication of the sub-parts is not making sense to me. How, to understand or get better mental model of it?
 A: The rule of multiplicatinn is simple.   If there are $x$ ways to perform a subtask and $y$ ways to perform a second (independent) subtask, and the tasks are performed in serial then there are $xy$ ways to perform whole task.
Think of it as branching paths.   If the path branches into $x$ paths and then each of these branches into $y$, how many ways are they to take?
Similarly if you take a bushwalk which has $x$ paths to get from $A$ to $B$ and there are $y$ paths from $B$ to $C$, then how many different ways are there to go from $A$ to $B$ to $C$?
When independent subtasks are performed in serial then multipling the counts of subtasks obtains the count of the whole task.
[Serial: both tasks must be performed; Parallel: the tasks are alternatives.]

We write $3!$ as the count of ways to arrange three distinct items; and likewise $7!$ and $10!$ are the ways to arrange seven, or ten, distinct items, respectively.
Well, now we write $\binom {10}3~$, or sometimes $~^{10}\mathbb C_3$, to represent the count of ways to choose three from ten distinct things.
But what exactly is this $\binom {10}{3}$ notation?   How do we find its value?
You have examined two ways to count how to arrange ten children among ten seats; when the seats are in two rows; three and seven.
Because these are ways to perform the exact same task; the counts must be the same.


*

*You have counted the ways to select three of ten children, then arrange them among three seats, and finally arrange the remaining children among seven seats.   This is: $~\binom{10}3\cdot 3!\cdot 7!$.

*You have also counted the ways to simply arrange all ten children among all ten seats.   That is: $~10!~$.
Hence we find that: $\dbinom{10}3 = \dfrac{10!}{3!\,7!}$ .

In general we find that $\dbinom{n}{r} ~=~ \dfrac{n!}{r!~(n-r)!}$ for all non-negative integer $r,n$ such that $0\leq r\leq n$
You use verify this with smaller examples: What are $\binom 3 1$ and $\binom 4 2$ ?
A: No matter if the bench is in the front row or in the back row, all 10 benches are distinctive. That's why the result is $10!$.
You can also think of it this way. We need to pick 3 kids to sit in the first 3 seats, and these 3 kids can be any 3 of the 10 kids. There are $\displaystyle \binom{10}{3}$ ways to do this. Then we don't have choice over who belongs in the other 7 benches, because there are only 7 other kids left.
For each option of which 3 kids sit in the front row, we have $7! \times 3!$ ways to seat the kids.
$\displaystyle \binom{10}{3}$ is the number of ways to group the 10 kids into a group of 3 and a group of 7. And $\displaystyle \binom{10}{3}$ would give the same result.
