Lattice pathsGrids up to 10
Count the routes across a grid, moving only right or up, and show why it is a combination, not a power of two.
Walk across a grid of streets from one corner to the far one, only ever going right or up, and count the different routes. It feels like a fresh choice at every corner — right or up, so two to the power of the steps — but it is really one choice made once: which of the equal-length steps go right.

The answer key prints on a separate page.
Every route across an m-by-n grid is exactly the same length: m steps right and n steps up, m + n steps in all. So a route is nothing more than a choice of WHICH of those steps are the rights — and that choice is a combination, (m + n) choose m. The move that makes it countable is seeing that the drawing is a choice in disguise: you never list a single path, you count the ways to place the rights among the steps.
The tempting wrong answer is 2 to the power of (m + n): at each step you go right or up, two choices, so surely two multiplied that many times. It over-counts wildly, because it allows more than m rights or more than n ups — routes that march straight off the edge of the grid and never arrive. The real count is far smaller, and the gap between the two is the whole lesson: free choices at every step are not the same as choices that have to add up to a fixed destination.