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.
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.