Number of strings that have a descent at the $k$th position of the string Let $X=\{1,2,3,\dots,n\}$. We make strings of length $n$ using elements from $X$. We call the string has a descent at $k$ if the number at the $k$th position is bigger than the number at the $k+1$ position.
Let $k$ be fixed and $1\leq k\leq n$. How many strings do we have that have only one descent and that descent is at $k$?
After trying some examples I guess the answer is $n-k$, yet I have no idea how to show it. Any help, please?
Update: Ok, now I have to find the number of strings that have at most one descent with $k$ unfixed. I think we first need to find the number of strings that have only one descent, namely
$${n\choose 1}-1+{n\choose 2}-1+\dots+{n\choose n}-1$$
Then we have to find the number of strings that have no descents. I think the string without descents can only be $(1,2,3,\dots,n-1,n)$.
Is my idea ok?
 A: Call your string $s_1,s_2,\ldots,s_n$.
$s_1,\ldots,s_{k}$ and $s_{k+1},\ldots,s_n$ should be increasing, moreover, $s_k > s_{k+1}$. Therefore $s_k \in \{k+1,k+2,\ldots,n\}$.
Also notice that for each fixed value for $s_k$, if we determine the set $\{s_1,s_2,\ldots,s_{k-1}\}$, then the rest of the string is uniquely determined: It will be the
sorted(${\{s_1,s_2,\ldots,s_{k-1}\}}$) , $s_k$ , sorted(${\{s_{k+1},s_{k+2},\ldots,s_{n}\}}$).
But if $s_k=i$ for $i=k+1,k+2,\ldots,n$ the number of possibilities for  the set $\{s_1,s_2,\ldots,s_{k-1}\}$ is simply $\binom{i-1}{k-1}$ therefore the solution is
$$\sum_{i=k+1}^n\binom{i-1}{k-1} = \binom{k}{k-1}+\binom{k+1}{k-1}+\ldots+\binom{n-1}{k-1} = \binom{n}{k}-1.$$

PS. Now I see there was a simpler argument.
Juse choose a $k$-subset of $[n]$ except for $\{1,2,\ldots,k\}$,and call it $A$. The strings you are looking are equivalents of
sorted($A$), sorted($A^c$)
A: If a string $s = (s_1, s_2,\dots, s_n)$ has only one descent at $k$ and each element is used once, then $1 \le s_1 < s_2 < \dots < s_k > s_{k+1} < s_{k+2} < \dots < s_n\le n$. It follows that $k < s_k \le n$.
For each $s_k \in \{ k+1,\dots, n\}$, the number of strings that satisfy the condition is ${s_k - 1}\choose{k-1}$.
Hence, the answer total number will be $\sum_{s_k = k+1}^n {{s_k - 1}\choose{k-1}} = {{k}\choose{k-1}} + {{k+1}\choose{k-1}} +\dots +{{n-1}\choose{k-1}} = \frac{{{n}\choose{k-1} }(n-k+1)}{k}-1 = {n \choose k} -1.$
Update: That's correct, there is only one arrangement if there is no descent. You can simplify the sum a little bit and be careful about the edge cases. $k$ can be at most $n-1$ and the total number of strings with at most $1$ descent is
$1+\sum_{k=1}^{n-1} \left( {n \choose k} -1\right) = 1+ 2^n -2 -(n-1) = 2^n-n.$
P.S. another way to interpret this is that for each integer in $\{1,2,\dots,n\}$, we are placing it either before the descent sign or after the sign, and once that is determined, there is only one way to arrange them. This would mean $2^n$ strings. However, certain subsets are invalid; if you choose $\{1,\dots,i\}$ as the subset before, for any $i\in \{1,\dots,n\}$, the resulting string does not have a descent. Subtracting these subsets, we are left with $2^n-n$.
