TITLE: Proving the number of permutations $A,B\;$ with $n+1$ total cycles and $AB=(123\cdots n)$ is $C_n$
QUESTION [10 upvotes]: Please give a combinatorial proof of the following:
The number of pairs $(A, B)$ of permutations of the set $\{ 1, 2,\ldots,n \}$ such that  they have a total of $n+1$ cycles and their composition $AB$ is the permutation $(123 \ldots n)$, is the $n$th catalan number.

REPLY [3 votes]: The desired product $(123\ldots n)$ increases $n-1$ elements and decreases only one. The number of elements it increases is at most the sum of the numbers of elements increased by each factor. Each $k$-cycle of a permutation increases at most $k-1$ elements. The sum of the lengths of all cycles in $A$ and $B$ together is $2n$. Since there are $n+1$ cycles in total, at most $2n-(n+1)=n-1$ elements can be increased. Since we want to increase $n-1$ elements, the $k$-cycles in $A$ and $B$ must each increase $k-1$ elements, that is, they must cyclically permute the elements in their orbit. Thus, each partition of $\{ 1, 2,\ldots,n \}$ yields at most one candidate for $B$, namely the permutation that cyclically permutes each block of the partition.
However, not all these candidates are suitable. In fact, precisely those permutations that cyclically permute each block of a noncrossing partition are suitable. To see this, consider first this representation of such a permutation $B$ (in red) together with the corresponding factor $A=(123\ldots n)B^{-1}$ (in green):
               
The fixed points are not shown. Note that all cycles run clockwise as they must. A noncrossing partition with $k$ blocks (corresponding to a permutation $B$ with $k$ cycles) partitions the circle into $n-k+1$ regions outside the blocks: There is one region to begin with, and each $k$-cycle splits one region up into $k$ pieces. As you can convince yourself from the illustration, the factor $A$, which has to move each element to the successor of its preimage under $B$, contains one cycle for each region outside the blocks of $B$. Thus the total number of cycles in $A$ and $B$ is $k+(n-k+1)=n+1$ as required. Note that $A$ also cyclically permutes each block of a noncrossing partition.
Now consider this representation of a permutation $B$ which cyclically permutes each block of a crossing partition (in red), again together with the corresponding factor $A=(123\ldots n)B^{-1}$ (in green):
               
Note that now the arrows of $A$ don't form counterclockwise cycles. To see that they can't, take any arrow $r$ of $B$ that crosses another arrow of $B$, say, $5\to0$, and find the arrow $s$ that crosses it closest to its origin, in this case $2\to8$. Viewed along $r$, $s$ must go left, since it is part of a clockwise cycle. (If it's part of a $2$-cycle, we can pick the arrow that goes left.) Now take the arrow $t$ of $A$ that gets composed with $s$, in this case $8\to3$. It crosses $r$ closer to the origin of $r$ than $s$ does, or it may end at the origin of $r$ (in this case $5$). Any clockwise cycle of $A$ containing $t$ would have to include an arrow $u$ of $A$ that either crosses $r$ yet closer to its origin, or begins at the origin of $r$. Then the arrow of $B$ that gets composed with $u$ would either cross $r$ yet closer to its origin, which is impossible by assumption, or end at the origin of $r$. The latter is also not possible because this arrow would be coming from the wrong side of $r$ to form a clockwise cycle with $r$.
In summary, the possible values of $B$ are precisely the permutations which cyclically permute each block of a noncrossing partition of $\{1,\dotsc,n\}$. It is well-known to those who know it that the noncrossing partitions of $\{1,\dotsc,n\}$ are counted by the Catalan numbers; a bijection with Dyck paths is exhibited here (section 3, page 5).