Why is the main operator involved with permutations and combinations the product operator? If I have $6$ people who happen to have the unfortunate names $A, B, C, D, E, F$ respectively, and I have $4$ seats for them to sit,
Why are the number of permutations of people in the first $2$ seats $6*5$? I get that for the first seat $6$ people may sit, and for the second $5$ may, as there are only $5$ left at this point, but why is it $6*5$ and not some other relation to find the number of arrangements for those first two seats? I can prove this happens to be the case by doing it manually, but I'm looking for a more intuitive look than that.
 A: Imagine making a table of permutations for each of the ways you can fill the first two seats. You have six rows, each labeled with someone's name,
and six columns, each labeled with someone's name.
In each cell, if we can fill the first seat with the person whose name is the label of the row and fill the second seat with the person whose name is the label of the column, we write the name of the first person followed by the name of the second person, and say this is how we're filling the first two seats.
But if we cannot do this, we leave the cell empty.
Each cell of the table gives us exactly one way to fill the first two seats,
except for the cells where the row label and column label are the same.
Those are the empty cells.
Given six people to be seated, we have six rows of the table, and in each of those rows there are six columns, but one column has a blank cell, so the row contains only five ways to fill the first two seats.
Six bags of objects, each bag contains five objects, how many objects are there?
That's the intuition. If you want a formal proof it can be done by induction on the number of people, at the cost of losing some of the intuitive simplicity of the tabular answer.
A: The process of seating $6$ individuals on $2$ seats can be broken into two parts:
$Part  \ 1:$ Selecting someone for the first seat
Clearly there are $6$ ways to do so. Say you selected someone and that person now occupies that seat.
$Part \ 2:$ Selecting someone for the second seat after the first seat has already been allotted
Only $5$ ways remain now that one person is already sitting on the first seat.

So, any choice for the first seat allows exactly $5$ possible cases. Since you have $6$ ways to choose who sits on the first seat, there are a total of $6\times 5$ possible cases.
