In how many ways in the set {1,2,..,n} the element $k$ is greater than all the numbers on its left? 
In how many ways in the set {1,2,..,$n$} the element $k$ is greater than
all the numbers on its left?
Example: $k=4$ $n=5$, 23415 is legal, 54312 is illegal

My attemp: $\sum_{i=0}^{n-k}\binom{k-i}{1}\left ( k-i-1 \right )!$
$\binom{k-i}{1}$ possible ways to find a place for $k$ such that its index can't be greater than $k$
and depending on it's current index there is $\left ( k-i-1 \right )!$ numbers left to arrange that can be smaller than $k$
Is my solution correct?
 A: By ignoring all elements other than $k$ and larger, we see that your original sequences will satisfy the property iff $k$ is the smallest of these.  The probability of this happening is $\dfrac{1}{n-k+1}$ so the total number of such sequences (including those elements smaller) is then $$\frac{n!}{n-k+1}$$
Alternatively, if you dislike using a probabilistic argument in this, approach by first picking the positions of the elements $k$ and greater.  The first of those must be occupied by $k$ but the remaining can be in any order.  Finally, those elements smaller than $k$ can be in any order in what remains, giving:
$$\binom{n}{n-k+1}(n-k)!(k-1)!$$
which of course simplifies again to the above.
A: If you want to think for each position from the left then here is how it will look -
Say $k$ is at the first place (from the left). You can permute the left of it in $^{(k-1)}P_0$ ways. You have to basically choose and arrange all the numbers lower than $k$ to its left. To the right you can permute it in (n-1)!.
Say $k$ is at the 2nd place. You can permute the left in $^{(k-1)}P_1$ ways and the right in $(n-2)!$ ways.
For $k$ in $i$ position, left to $k$ will be $^{(k-1)}P_{(i-1)}$ permutations and to the right it will be (n-i)!
So, $\sum \limits_{i=1}^{k} {^{(k-1)}P_{(i-1)}} (n-i)!$ is what you are looking for.
