The Catalan number $C_n= \frac{2n \choose n}{n+1}$ counts the number of binary strings with $n$ $1$s and $n$ $0$s so that every prefix has at least as many $1$s as $0$s. There is a combinatorial proof, I think due to Cauchy, which involves an 18th camel.

There are $2n \choose n$ binary strings with $n$ of each letter. Add an $n+1$st $1$ to the end of each word! There are $n+1$ cyclic rotations ending in $1$. Precisely one of these has the property that each prefix has at least as many $1$s as $0$s. (This is nontrivial but not hard to prove.)

For example, the $5$ rotations of $11100010$ are

$$\begin{eqnarray}11100010{\color{red}1} \newline 11000101{\color{red}1} \newline 10001011{\color{red}1} \newline 00010111{\color{red}1}\newline 01111000{\color{red}1}\end{eqnarray}$$

with the 18th camels in red.