4. A rectangle with dimensions $m$ and $n$, where $m$ and $n$ are natural numbers and $n = m k$, is divided into $m \cdot n$ unit squares. Each path from point $A$ to point $C$ along the divided segments (sides of the small squares) where movement to the right and movement upwards is allowed, has a length of $m+n$. Find how many times the number of paths from $A$ to $C$ that pass through point $T$ is greater than the number of paths from $A$ to $C$ that pass through point $S$.
Solution. We will denote every path from $A$ to $C$ with zeros and ones, where a one represents a movement upwards by one unit of length, and a zero represents a movement to the right by one unit of length. According to this, the number of paths through point $T$ is

$$
C_{m+n-1}^{n-1}=\binom{m+n-1}{n-1}
$$

and the number of paths through point $S$ is

$$
C_{m+n-1}^{m-1}=\binom{m+n-1}{m-1}
$$

Since

$$
\frac{C_{m+n-1}^{m-1}}{C_{m+n-1}^{n-1}}=\frac{\binom{m+n-1}{m-1}}{\binom{m+n-1}{n-1}}=k
$$

it follows that the number of paths through point $T$ is $k$ times greater than the number of paths through point $S$.