An ant starts from the origin of a coordinate system, $(0, 0)$, and in each step moves either up or right by one unit. After $10$ such moves, what is the probability that the ant settles down at point $(4, 6)$?
I understand that it's a partition problem. Because probability for either up or right is $1/2$ and the ant has to make $6$ moves up and $4$ to the right, would it be ${10 \choose 4} \left(\frac{1}{2} \right)^4 \left(\frac{1}{2} \right)^6 = {10 \choose 4} \left( \frac{1}{2} \right)^{10} = 0.2051$?
Yep. And here's a figure that may help you visualize your solution.