Willy Walks to School (Part I)

Let us think of each movement through the roads as a pace . That is, the movement from $(x, y)$ to $(x + 1, y)$ or from $(x, y)$ to $(x, y + 1)$ is one pace. Since it takes $10$ paces to travel from the origin to $(10, 0)$ and $12$ paces to travel from the origin to $(0, 12)$ , travelling to $(10, 12)$ takes $22$ paces.

Considering that Willy is to walk strictly either right or up to go to school in the most efficient way, then, we can think of those twenty-two paces as a combination of binary movements that either go up or right. Now, if Willy has to advance strictly ten paces to the right (and the rest are 'up' paces), there are ${22 \choose 10} = 646 646$ ways to do so. ( NOTE : You will arrive at the same result if you considered the twelve 'up' paces instead of the ten 'right' paces: ${22 \choose 12} = 646 646$ .)

Hey William Sorry to interrupt in another question but I think your answer for willy ....part 2 is wrong... Thinking in the same way...answer should be 12C6*3C2....is something wrong in here????