How many pairs of dance partners can be selected from a group of $12$ women and $20$ men? How many pairs of dance partners can be selected from a group of $12$ women and $20$ men ?

Ans given : $P(20, 12)$

Shouldn't the answer be $20 × 12$ as the pair can be selected from any of the $12$ women and for each women there are $20$ men to choose for.
You first select 12 men from possible 20, that can be done in $\binom{20}{12}$ ways. Now these 12 men have to be paired with the 12 women. Each pairing is simply a bijective function from the set of 12 men to the set of 12 women. Number of such bijective mappings is $12!$. So in all
$$\binom{20}{12} \cdot 12!=P(20,12) \quad \text{ways}.$$
Let $W$ denote the set of women and $M$ denote the set of men. An arrangement in which each woman has exactly one male dance partner is the same thing as an injection $W \rightarrow M$. So compute:

$$|M^{\downarrow W}| = |M|^{\downarrow |W|} = \prod_{w = 0}^{|W| - 1}(|M|-w).$$

Notation. whenever $A$ and $B$ are sets, the notation $A^{\downarrow B}$ means the set of all injections $B \rightarrow A,$ and whenever $a$ and $b$ are natural numbers, $a^{\downarrow b}$ denotes the falling factorial. Now just plug your numbers in.