Number of permutations in which a integer appears before other Two integers $x$, and $y$ are fixed such that $1 \le x &lt; y \le n$.
I am trying to count the number of permutations of {$1,2,3,4,...,n$}
($n \ge 2$) were $x$ appears to the left of $y$. 

I think I should use product rule in this case but I am not sure.
Hint: 

To answer how many of these will have $x$ to the left of $y$,

think about how many of these will have $x$ to the right of $y$.

Are these numbers related?


  There are exactly as many permutations with $x$ to the left of $y$ as there are having $x$ to the right of $y$.  See if you can explain why using a bijective function between them.


What does this imply about the total number of permutations in relation to the number of permutations with $x$ to the left of $y$?



For a direct counting argument instead of a combinatorial argument, first pick which two spaces the pair $x,y$ will occupy.  Whichever two spaces you picked, put $x$ in the left space and $y$ in the right.  From there, fill in the remaining spaces however you prefer (left to right or smallest to largest, etc...).


  There are then $\binom{n}{2}$ ways of picking which two spaces $x$ and $y$ collectively occupy and $(n-2)!$ number of ways of picking the arrangement of the remaining numbers.  Apply multiplication rule.
Take the position of $y$ and make this partition of the set:

$$\{1,2,...,n\}=\{1,...,i\}\cup\{y\}\cup\{j,...,n\}$$

Now count when $x \in\{1,...,i\}$.

This is the canonical way to solve this kind of permutations, how it appear in textbooks (by example Bijective combinatorics if I remember correctly). 

It works for a lot of words with order restrictions. In this case it is very easy cause there is only one restriction, i.e. $x&lt;y$ and the answer is "trivialy" $n!/2$ (just by symmetry).