What does $^{6}P_{3}$ mean here? Can anyone explain me this statement?

"$3$ people can occupy $3$ places out of $6$ places in $^{6}P_{3}$ ways!" 

How? I thought $^{6}P_{3}$ was supposed to mean arranging $6$ objects at $3$ places?
Also any tips on how to improve my understanding towards combinatorics? Thanks, I've started out, and it's clearly not easy as other math topics. So any tips would be nice... Thanks again!
 A: ${}^6\mathrm P_3$ means simply $6\times 5\times 4$ - that is, the product of $3$ terms starting from $6$ and decreasing by $1$ each time.
This is equal to the number of ways to arrange $6$ objects into $3$ places - you have $6$ options for the first place, $5$ options for the second place and $4$ options for the third. But it could also be the number of ways to arrange three people/objects into six places - you have $6$ choices for where to put Alice, then $5$ choices for Bob and $4$ for Charlotte.
It's quite common that you can think about a problem involving objects and places either way round; sometimes it is easier to think about how many options there are for each place, and other times about options for each object.
A: The formula for $^nP_r$ is $$\frac{n!}{(n-r)!}$$ which will always simplify down to $$n(n-1)(n-2)\cdots(n-r+1)$$
For your problem, we have $3$ people and we have $6$ available spots to put these people in. Only one person can occupy a single spot at a time. So the first person has $6$ spots to choose from, the second has $5$ spots, and the last guy has $4$. This means the total number of arrangements is $$6\cdot 5\cdot 4 = \ ^6P_3$$
Notice that the order matters in this situation, as even if we were to arrange the three people in seats 1,3, and 5, it matters which person sits where. If instead we didn't care which person sat where and we were only concerned with the number of ways we could choose the chairs being used, then the answer from before would need to be divided by $3!$. This is what is known as a combination, rather than a permutation. 
In general if we have $n$ chairs and we want to sit $r$ people at these chairs in a way that we don't care who sits in which chair, but we want to know the different combinations of chairs being used, then the answer is $$\frac{n!}{r!(n-r)!} = \frac{n(n-1)(n-2)\cdots(n-r+1)}{r!} =\binom{n}{r} = \ ^nC_r$$
