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Let $G$ be a simple directed graph with vertex set $\{(i,j) \in \mathbb{Z}^2 : 15 \leq i \leq 21 , 21 \leq j \leq 25\}$ and an arc from $(i,j)$ to $(i',j')$ if and only if $(i',j')$ is obtained from $(i,j)$ by adding 1 to one coordinate. Then the number of directed paths from $(15,21)$ to $(21, 25)$ is

(a) $1260$

(b) $1420$

(c) $980$

(d) None of these

My answer: The number of paths from $(15,21)$ to $(21, 25)$ would be $${(21-15)+(25-21) \choose (25-21)} = {10 \choose 4} = 210.$$

Is this correct? Just want to get a second opinion.

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Miz
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Yes. Your answer is correct. The solution to the general problem is that you must take X right steps, and Y down steps then the number of paths is simply the ways of choosing where to take the up (or right) step. i.e.

((X+Y),X)=((X+Y),Y)

For example, you are traversing graph points of equal distance i.e. squares then there are 6 right steps and 4 up step :

((6+4),4) = ((6+4),6) = (10,4) = (10,6) = 210

P.S (a,b) -> a! / b!

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