The [OEIS entry for Pascal’s triangle](https://oeis.org/A007318) contains the following intriguing remark:



> 
> $C(n,k)$ = the number of Dyck paths of semilength $n$, with $k$ "u"'s in odd numbered positions and $k$ returns to the x-axis.
> 
> 
> Example: {`U` = u in odd position and `_` = return to x axis}
> 
> 
> $\mathrm{binomial}(4,1)=1$ (`Uudududd_`)  
> 
>  $\mathrm{binomial}(4,2)=3$ (`Uududd_Ud_`, `Ud_Uududd_`, `Uudd_Uudd_`)  
> 
>  $\mathrm{binomial}(4,3)=3$ (`Ud_Ud_Uudd_`, `Uudd_Ud_Ud_`, `Ud_Uudd_Ud_`)  
> 
>  $\mathrm{binomial}(4,4)=1$ (`Ud_Ud_Ud_Ud_`)
> 
> 
> – Roger Ford, Nov 05 2014
> 
> 
> 


It isn’t obvious to me why this should be true. I found the edit where this remark was added, and it doesn’t include any links or references.


Can anyone give a proof – or supply a reference, if this is a known result?


**Added**: As Anthony Quas points out in the comments, the example doesn’t match the description. Based on the example the claim should presumably be that $n \choose k$ is the number of paths of semilength $n+1$ that have $k+1$ u’s in odd positions and $k+1$ returns. ~~But this error makes me wonder if the whole thing is nonsense. I’ll check some more examples.~~



Counting particular Dyck paths
The answer is pretty simple. Consider any of $k$ parts between consecutive touches of $x$-axis. Note that we can assign a unique `U` character to each of these parts, namely, the first character, which meets our quota of `U` characters. The last character of each part obviously has to be `d`. Between the first and the last character of a part, no character at odd position can be `U` (since the quota is met), so they have to be `d`. All the other characters in this part have to be `u` to compensate the balance, so the only way a part can look like is `Uudud...udd` for some number of repetitions of `ud`. So the answer is equal to the number of ways to cut a sequence of length $n$ into $k$ non-empty parts, which is ${n - 1 \choose k - 1}$.


