Consider an equilateral triangle of side length $n$, which is divided into unit triangles, as shown. Let $f(n)$ be the number of paths from the triangle in the top row to the middle triangle in the bottom row, such that adjacent triangles in our path share a common edge and the path never travels up (from a lower row to a higher row) or revisits a triangle. An example of one such path is illustrated below for $n=5$. Determine the value of $f(2005)$.

![](https://cdn.mathpix.com/cropped/2024_04_17_af19c08dc8a5848f069fg-1.jpg?height=325&width=371&top_left_y=824&top_left_x=926)
We shall show that \( f(n) = (n-1)! \).

Label the horizontal line segments in the triangle \( l_{1}, l_{2}, \ldots \) as in the diagram below. Since the path goes from the top triangle to a triangle in the bottom row and never travels up, the path must cross each of \( l_{1}, l_{2}, \ldots, l_{n-1} \) exactly once. The diagonal lines in the triangle divide \( l_{k} \) into \( k \) unit line segments and the path must cross exactly one of these \( k \) segments for each \( k \). (In the diagram below, these line segments have been highlighted.) The path is completely determined by the set of \( n-1 \) line segments which are crossed. So as the path moves from the \( k \)th row to the \( (k+1) \)st row, there are \( k \) possible line segments where the path could cross \( l_{k} \). Since there are \( 1 \cdot 2 \cdot 3 \cdots (n-1) = (n-1)! \) ways that the path could cross the \( n-1 \) horizontal lines, and each one corresponds to a unique path, we get \( f(n) = (n-1)! \).

Therefore \( f(2005) = (2004)! \).

![](https://cdn.mathpix.com/cropped/2024_04_17_af19c08dc8a5848f069fg-1.jpg?height=322&width=384&top_left_y=1975&top_left_x=903)