How many ways can 2 person sit in 4 empty chairs? I can find the answer using brute force as 12, but what is the formula to calculate this for any combination of person and chairs.
Here is the brute force combinations for 2 person, 4 chair:
Group where A is always placed before B

A,-,-,B,
A,B,-,-
-,-,A,B
-,A,-,B
A,-,B,-
-,A,B,-

Group where B is always placed before A

B,-,-,A
B,A,-,-
-,-,B,A
-,B,-,A
B,-,A,-
-,B,A,-

 A: Seat A first, then B.  A has $4$ choices, leaving $3$ choices for B, giving a (multiplicative) total of $4\times3=12$ different seatings.
A: Choose 2 seats out of 4 for the two people and the 2 people can arrange themselves in $2!$ ways. Thus the answer is
$$2! \times \binom{4}{2} = 2 \times 6 = 12$$
For $n$ chairs and $m$ people (assuming $\binom{n}{m} = 0$ for $m \ge n$) this reduces to choosing $m$ seats out of $n$ and then permuting the $m$ people which is given by the formula
$$m! \times \binom{n}{m}$$
Here $\binom{n}{m}$ is the binomial coefficient which denotes the number of ways to choose $m$ objects from a collection of $n$ distinct objects.
The number $m! \times \binom{n}{m}$ is also denoted as $^nP_m$.
A: Some would call this the "fundamental principle of counting"; multiply the options at each step, e.g., in this case, $4 \times 3 = 12$. 

In combinatorics, the rule of product or multiplication
  principle is a basic counting principle (a.k.a. the fundamental
  principle of counting). Stated simply, it is the idea that if there
  are $a$ ways of doing something and $b$ ways of doing another thing,
  then there are $a · b$ ways of performing both actions.

Rule of product, Wikipedia
A: The formal answers are addressing the two following common sense situations: 
(1)  The 1st person can sit in any of 4 chairs.
For each of the 4 possible choices the st person makes there are 3 left to the 2nd person, so there are 4 sets of 3 combinations.
4 x 3 = 12 
(2) As before the 1st person can sit in any of 4 chairs.
Now seat the 2nd person without regard to the location of the 1st person.
The 2nd person can ALSO sit in any of 4 chairs for each choice made by the 1st person.
4 x 4 = 16.
BUT of these 16 combinations, 4 have both people in the same chair.
If this is not intended then these 4 combinations are invalid
so 16-4 = 12. 
(3) :-)
If anyone sits anywhere there are no longer 4 empty chairs, so there are 0 ways that 2 people can sit in 4 empty chairs.
