Number of sequences with $k$ pairs of consecutive (+1)(-1) I'm working on a combinatorics problem related to Dyck paths/Ballot sequences/Catalan numbers and the Narayana numbers (defined by $N(n,k) = \frac{1}{n} \binom{n}{k}\binom{n}{k-1}$).
My question is this: Let $P$ be the set of sequences of $\{1^{n+1},(-1)^n\}$, i.e. consisting of $(n+1)$ positive ones and $n$ negative ones, which start with $1$ and contain exactly $k$ pairs of consecutive $(+1)(-1)$s. Prove that $|P| = \binom{n}{k-1} \binom{n-1}{k-1}$.
So I essentially have a binary word, beginning with a $1$, of length $2n+1$ with $k$ pairs of $+1$ followed by $-1$. I've tried to enumerate this by choosing where to put the pairs, but I am having trouble since the second position is special (it comes after a $1$, so if the second position is a $-1$, this adds a pair).
 A: For convenience of notation let me replace $-1$ with $0$. Any string consists of alternating blocks of $1$s and $0$s, beginning with a block of $1$s, and it’s not hard to see that a string has $k$ $10$ pairs if and only if it has exactly $k$ blocks of $0$s. Ignore the leading $1$. A standard stars and bars calculation shows that there are $\binom{n+1}k$ ways to distribute $n$ $1$s amongst the $k+1$ slots before the first block of $0$, between adjacent blocks of $0$s, and after the last block of $0$s under the requirement that the middle $k-1$ slots be non-empty. Similarly, there are $\binom{n-1}{k-1}$ ways to distribute the $n$ $0$s amongst the $k$ blocks of $0$s. Thus, there should be $\binom{n+1}k\binom{n-1}{k-1}$ sequences with $k$ $10$ pairs. Note that
$$\begin{align*}\binom{n+1}k\binom{n-1}{k-1}&=\frac{n+1}{n+1-k}\binom{n}k\binom{n-1}{k-1}\\
&=\frac{n+1}{n+1-k}\binom{n}k\binom{n}{k-1}\frac{n-k+1}{n}\\
&=(n+1)N(n,k)\;.
\end{align*}$$
$$$$
A: This is not provable because it is not true. The question was supposed to also have the constraint of ending on $(-1)$.
