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The nth Catalan number, $C(n)$ counts the number of binary strings with n $0$'s and n $1$'s such that any initial substring has at least as many $0$'s as $1$'s.

I know that the formula for the nth Catalan number is $C(n)=\frac{1}{n+1}\binom{2n}{n}$ and the formula for the number of binary strings with n $0$'s and n $1$'s is just $B(n)=\binom{2n}{n}$

Is there an natural way to partition the set of all binary strings with n $0$'s and n $1$'s into sets of size $n+1$ such that each partition contains exactly one string with the property that any initial substring contains at least as many $0$'s as $1$'s?

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The answer is yes; it’s implicit in this proof that $C_n=\frac1{n+1}\binom{2n}n$. In the terminology there, start with any path whose exceedance is $0$, and reverse the algorithm to produce in succession paths with exceedance $k$ for $k=1,\dots,n$.

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I've stolen the following quote from Four Proofs of the Ballot Theorem by Marc Renault in Mathematics Magazine, Volume 80, No. 5, December 2007. The whole article makes very interesting reading.

Consider the ${2n\choose n}$ possible lattice paths starting from the origin and consisting of $n$ upsteps $(1,1)$ and $n$ downsteps $(1,-1)$. It turns out, surprisingly, that the number of these paths with $i$ upsteps above the $x$-axis $(0\leq i\leq n)$ is the same, regardless of the value of $i$. Consequently, the number of paths with all $n$ upsteps above the $x$-axis must be ${2n\choose n}/(n+1)$.

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