Number of ways to arrange couples on a bench I'm trying to solve the following question:

We have $n$ couples of men and women. We can assume that $n$ is odd. We need to arrange the people on a bench, so that at least half of the women sit on the right side of their spouses (not necessarily right after them). What is the number of ways to do so?

My book states that the solution is:
$$
\frac{(2n)!}{2}
$$
But I can't seem to understand why. Does it have something to do with $n$ being odd? What if $n$ is even?
 A: Hint : What would be the answer to the following question (no need to read it, I just copy-pasted it and change right to left) :

We have n couples of men and women. We can assume that n is odd. We need to arrange the people on a bench, so that at least half of the women sit on the left side of their spouses (not necessarily right after them). What is the number of ways to do so?

More aggressive hint

 The answer to both question is the same, also in any arrangement of all $2n$ people, either half of the women sit on the right side of their spouses or half of the women sit on the left side of their spouse. Both are disjoint arrangement (this is where you use the fact that $n$ is odd), therefore it is $\frac{(2n)!}{2}$

A: There are $n$ men and $n$ women, giving a total of $2n$ people. The number of ways to arrange them is $(2n)!$. In half of these arrangements, at least half of the women sit on the right side of their spouses, and in other half, at least half of the women sit on the left side of their spouses (this is because of symmetry of the arrangements$^*$).
So, the number of required combinations is
$$\frac{(2n)!}2$$
$^*$Think of it this way- if you hold a mirror on one side of any arrangement, the image you will see will be the same except with reversed parity. Note that the fact that $n$ being odd guarantees that the arrangements of the actual setting and the arrangements of the image are all disjoint.
A: Here's a more direct approach.
The number of ways to place all $2n$ individuals on the bench in such a way so that exactly $k$ of these women are seated on the right of their partners and the other $n-k$ women are seated on the left of their partners is ${2n \choose 2}{2n-2 \choose 2}\dots {2\choose 2}=\frac{(2n)!}{2^n}$. Note this figure doesn't dependent on $k$. Because there are ${n \choose k}$ ways to designate the $k$ women who are seated on the right of their partners, the desired number of seating arrangements is $$\frac{(2n)!}{2^n}\sum_{k=\frac{n+1}{2}}^n{n \choose k}$$ Using the facts that $\sum_{k=0}^{\frac{n-1}{2}}{n \choose k}=\sum_{k=\frac{n+1}{2}}^n {n \choose k}$ and $\sum_{k=0}^n{n \choose k}=2^n$ we get that $\sum_{k=\frac{n+1}{2}}^n{n \choose k}=2^{n-1}$ which simplifies our answer to $\frac{(2n)!}{2^n} \cdot 2^{n-1}=\frac{(2n)!}{2}$ as required.
