Catalan numbers

Total. Number of paths with $n$ up-steps and $n$ down-steps: $\binom{2n}{n}$.

Bad paths. Reflect the initial segment up to the first dip below the axis: bijection with paths from $(0,-1)$ to $(2n,0)$ starting with $n+1$ up and $n-1$ down. Count: $\binom{2n}{n+1}$.

Conclusion. $C_n=\binom{2n}{n}-\binom{2n}{n+1}=\frac{1}{n+1}\binom{2n}{n}$.

Example. $C_3=\frac{1}{4}\binom{6}{3}=\frac{20}{4}=5$.