In the field mathematicus there is an infinite alignment of flowers.

A grasshopper moves from flower to flower. At each jump, it can land on the adjacent flower (if it is on flower $n$ it can jump to flower $n-1$ or $n+1$). Initially, the grasshopper is on flower 0.

In how many ways can it jump $k$ times, knowing that after its $k$ jumps, it must return to flower 0?

In how many ways can it jump $k$ times, knowing that after its $k$ jumps, it must return to flower 0 without ever having landed on a strictly negative flower?
For the grasshopper to be on flower 0 after $k$ jumps, there must have been as many jumps to the left as to the right, so it is necessary that $k$ be even, otherwise it is impossible. Let's denote $k=2r$.

It is necessary and sufficient that there be as many jumps to the right as to the left, so there are $\binom{2r}{r}$ ways to make $2r$ jumps for the grasshopper to return to its starting point. We could generalize the result by asking that the grasshopper be on flower $i$ after its $k$ jumps (how?).

If we now impose that the grasshopper never passes over a strictly negative flower, we easily notice that for all well-parenthesized words, there corresponds a path of the grasshopper by transforming a "(" into a jump to the right and a ")" into a jump to the left. Conversely, a series of $2r$ jumps ending on flower 0 corresponds to a unique well-parenthesized word. Thus, there are $C_{r}$ series of jumps.