One thousand exercise in probability third edition section 1.7 question 6 The question is following:
A group of $2b$ friends meet for a bridge soiree. There are $m$ men and $2b − m$ women where
$2 ≤ m ≤ b$. The group divides into $b$ teams of pairs, formed uniformly at random. What is the
probability that no pair comprises $2$ men?
The author give out answer as following:
There are ${2b \choose m}$ equal likely ways of assinging the men to the pairs. The number of assignations with no men pair is $2^m{b \choose m}/{2b \choose m}$
What confused me:
Firstly, how does the author derives the number of ways of assigning the men to the pairs is ${2b \choose m}$? By using a simple test, it is seems incorrect, for example, say b = 2 and m = 2, then according the answer, we would have ${4 \choose 2}$ ways of assigning the men to pairs which is 6. Let $m_{a}$ and $m_{b}$ represent the male a and male b, $f_{c}$ and $f_{d}$ represent female c and female d, then all the possible ways we can assign pairs are: (1).($m_{a}$, $m_{b}$) and ($f_{c}$, $f_{d}$) (2).($m_{a}$, $f_{c}$) and ($m_{b}$, $f_{d}$) (3). ($m_{a}$, $f_{d}$) and ($m_{b}$, $f_{c}$) which in total are 3 ways instead of 6 ways.
Therefore, I would like to know whether the sample answer is correct and if it is correct how to derive this answer.
 A: Yes it is correct. The author considers the pairs to be ordered and  they don't care about the identity of the persons, just their gender.
Think of all the persons arranged into a linear line and the pairs are formed by who are next to each other (i.e. 1 and 2 is a pair, 3 and 4 and so on). Then $ 2b \choose m$ is the number of ways to choose the places for all $m$ men in this line (but you don't care where Jack or Mark goes). The number ${b \choose m} 2^m$ is the way to choose the places for the men with no man-pairs: first choose the pairs where to place a man (this gives ${b \choose m}$) and then to which place of the two the man is placed ($2^m$).
In your example there are the pairs (you only put $m$ or $f$ no identity for persons is given)

*

*$(m, m)$ and $(f, f)$

*$(f, f)$ and $(m, m)$

*$(m, f)$ and $(m, f)$

*$(f, m)$ and $(m, f)$

*$(m, f)$ and $(f, m)$

*$(f, m)$ and $(f, m)$
A: The answer proposed, interpreted as a probability rather than as a number of assignations, is correct, but the explanation is greatly lacking. First of all the author should be precise about the method. The most natural thing to do here is to take as set from which an element is uniformly drawn the set of partitions of $2b$ elements into pairs, which set has $\frac{(2b)!}{2^bb!}=\prod_{i=1}^b(2i-1)$ elements, and this is not at all the denominator $\binom{2b}m$ used in the given formula.
Therefore another method than a uniform selection from this set is being used. Instead a uniform selection from all $(2b)!$ permutations is assumed (which is not that denominator either, but at least it is a multiple of it), which can be visualised as seating the $2b$ people in a bus with $b$ rows of $2$ places each. (In the end we don't care about permutations of the rows nor of permutations within each row, which is why the number of pairings is a quotient of the $(2b)!$ seat assignments; however uniformly choosing a permutation leads to uniformly choosing a pairing.) Now the trick that is silently used, is to first seat the $m$ men, after which it is already clear whether any pair of men is formed; for a probability computation one can ignore the $(2b-m)!$ remaining assignments of the women, since this number is the same irrespective of what assignment is given for the men. But given the set of seats assigned to the $m$ men we can also already determine whether any pair of men is formed, so we can also divide by the $m!$ permutations of the sets in that set. In the end we just choose $m$ seats to be occupied by men among the $2b$ available seats, and choosing an assignment uniformly leads to a uniform choice among these $\binom{2b}m$ possibilities.
So now we are down to counting among the $\binom{2b}m$ subsets of seats, how many avoid ever having both seats of a same row (even though such a subset does not at all determine an actual pairing of the $2b$ friends). The number of those choices can be found as the product of the number $\binom bm$ of choosing $m$ rows, and the number $2^m$ of ways of choosing one seat to be occupied by a man among in each of the chosen rows. Thus one ends up with the given probability of $\binom mb2^m/\binom{2b}m$, which is the correct answer.
My sincere feminist apologies for focussing on the men while they are the minority here. But this masculine centred approach is present already in the question and the proposed answer.
