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$$\displaystyle a(n,r) = a(n,r-1) + a(n-1,r)$$

for $n,r >1$. Given that $a(1,0) = a(0,1) = 1$.

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    hold on, what about $a(0,2)$ is it defned? – 2017-01-10

1 Answers 1

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Clearly $a(n,r)$ is the number of ways to get from $(0,0)$ to $(n,r)$ by only moving one unit up or right on the lattice plain. So we have $a(n,r)=\binom{n+r}{n}$

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    How would you derive such a relation? – 2017-01-10
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    well, the recurrences are the same. – 2017-01-10
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    Yes, I know that but I mean how would you do this systematically? – 2017-01-10
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    I'm sorry, I don't have experience solving $2$-dimensional recurrences systematically. You mean something like a g.f approach? – 2017-01-10
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    Nevermind, I found out how to solve this algorithmically. Btw nice intuition. – 2017-01-10
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    @MrYouMath does your german have an accent? I learned german a couple of years back and I wasnt sure. – 2017-01-10
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/51620/discussion-between-mryoumath-and-jorge-fernandez-hidalgo). – 2017-01-11