How many ways can 2 person sit in 4 empty chairs?

Seat A first, then B. A has $4$ choices, leaving $3$ choices for B, giving a (multiplicative) total of $4\times3=12$ different seatings.

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 $^nP_m$.

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