Number of ways of dividing 20 persons into 10 couples Here the first thing that we do is we choose 2 people from 20 then 2 from the remaining 18 then 2 from the remaining 16 ..... till 2 from the remaining 2 in 
$[\binom{20}{2}][\binom{18}{2}][\binom{16}{2}].....[\binom{14}{2}]$ ways.
i.e. $\frac{20!}{2^{10}}$ ways. However we divide the final answer by $10!$ because ‘order does not matter’ 
What does ‘order not matter’ mean in this context ? 
 A: Arrange the $20$ people in a line and pair the first two people up, the second two people up and so on to get something like
$$
(12)(34)\dotsb(19\,20).
$$
There are $20!$ orderings and each yields a way to pair the people as described above. But notice that every way to pair the people is generated by $2^{10}\cdot10!$ orderings since we can change the order of the pairs and the order within a pair without changing the collection of couples. Hence there are
$$
\frac{20!}{2^{10}\cdot10!}
$$
ways.
A: The first person can be coupled with $19$ possible people, the next person can be couple with $17
$ ... and so on . So there are $19!!$ possible ways. $ \frac{20!}{2^{10} 10!}= \color{blue}{654729075}$.
A: Here's a simplified version of the problem to illustrate the meaning of "order does not matter."  Suppose we had only four people $\{A,B,C,D\}$ and we needed to couple them.


*

*Couple 1: $A$ and $B$, Couple 2: $C$ and $D$

*Couple 1: $C$ and $D$, Couple 2: $A$ and $B$


Your process of choosing two among four, then two more among the remaining two, would treat these two couplings as different.  But the problem only asks to divide the people into (indistinct) couples, so these should be treated the same.
A: If you find yourself only having a calculator without a factorial function, just multiply the odd numbers up to 20:
(1)(3)...(17)(19) = 654729075
This is because the first person can be paired with 19 people, the second person can be paired with 17 people, etc. If n is the total number of objects and r is the number of objects grouped at a time, then you can use the formula:
$\prod_{i=1}^{n/r} ri-1$
In your case:
$\prod_{i=1}^{10} 2i-1$
