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.
 A: 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
$$\binom{20}{12} \cdot 12!=P(20,12) \quad \text{ways}.$$
A: 
Shouldn't the answer be 20 x 12 as the pair can be selected from any of the 12 women and for each women there are 20 men to choose for.

$20{\times}12$ counts the ways to select just $1$ m:f-pair from the group of $20$ males and $12$ females.
${}^{20}\mathrm P_{12}$ counts the ways to arrange the group into $12$ m:f-pairs (and $8$ male wallflowers).   All at once.
So the answer depends on how you interpreted the question.
A: There are n*(n-1)*(n-2)....(n-r+1) possibilities
(ie) 20 ways to select first pair
19 ways to select second pair
18 ways to select third pair
..........
9 ways to select last pair ie (20-12+1)
Thus 20*19*18....*9 ways are there
which is nothing but P(20,12)
A: 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.
