Suppose we have a set $S=\{1,2,3,x,y\}$. There are $5!$ ways to rearrange the elements in the set, but I am confused about how to find the number of ways to rearrange the set given that $3$ comes before $2$ which comes before $1$ (or something like that). For instance, we could write
$$\{3,2,1,x,y\}\\ \{3,2,x,1,y\}\\ \{3,2,x,y,1\} \ \text{etc}...$$
It is easy to do by a brute force method, consisting of writing down all the possible sets, but is there a short, mathematical, way to do so?
Given the condition that 3 comes before 2 which comes before 1, out of those $5!$ arrangements, the arrangements of $\{1,2,3\}$ within themselves have to be according to the condition.
So, out of the $3!$ possibilities of $\{1,2,3\}$ arranging within themselves, the only one that satisfies the given condition is $\{3,2,1\}$, that is one out of $3!$.
Hence, total number of arrangements is $\frac{5!}{3!}$.
To make things clearer and simpler, start by considering $\{1,2,3,4,5\}$ as a starting set
Also notice that you can just as well count permutations with $ 1' < 2' < 3'$ (I hope the notation is intuitive enough).
Then just count how many positions are valid for $1,2,3$, leaving $4,5$ aside. You can "count on your fingers" or see that there is some inductive formula to count this. You should get $10$.
Then for every valid positioning of $1,2,3$, you have $2!=2$ ways of positioning $4,5$.
So in your example, answer should be $20$.
