How many paths are there on an $n \times n$ grid to go from the point $(0 ; 0)$ to the point $(n ; n)$ using only moves of 1 up or 1 to the right?
This amounts to creating a word of $2n$ letters containing $n$ letters $D$ and $n$ letters $H$. There are $\left( \begin{array}{c} 2n \\ n \end{array} \right)$ ways to choose $n$ positions for the letters $D$, hence $\binom{2n}{n}$ possible paths.

Theorem 14. For $n \in N, 0 \leq k \leq n$,

$$
\binom{n}{k}=\frac{n!}{k!(n-k)!}
$$

Proof. An ordered list of $k$ elements chosen from $n$ can be formed by choosing the first element from $n$ ( $n$ possible choices), then the second element from $n-1$ ( $n-1$ possible choices), and so on, with the last element being chosen from $n-k+1$ elements. Therefore, there are $n(n-1) \ldots (n-k+1) = \frac{n!}{(n-k)!}$ ordered lists of $k$ elements chosen from $n$. However, one can also first choose the subset of $k$ elements from $n$ ( $\binom{n}{k}$ possible choices) and then order the set to form a list ( $k!$ possible orders). Therefore, there are $\binom{n}{k} \times k!$ ordered lists of $k$ elements chosen from $n$. By combining these two results, the formula is established.