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First of all this is not a duplicate of Proving Pascal's Rule .

I tried google before posting this question ,but did not find anything useful.

Here is what I already know : To get to any point the number of paths is just the number entering horizontally from the left plus the number entering vertically from below.

Let's say P(n,r) is the point I want to get to.

So, lattice method gives C(n,r) = C(n-1,r) +C(n-1,r-1)

But I think that C(n,r) = C(n-1,r) +C(n,r-1) and not C(n,r) = C(n-1,r) +C(n-1,r-1)

Because those are precisely the co-ordinates of the point one to the left, and the other immediately under.

What's wrong with my understanding?
Can someone show the lattice proof here or point me to the right place?

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mac07
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1 Answers1

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You did not count the number of moves correctly.

Suppose that you wish to move from $(0, 0)$ to $(n, r)$, where $n, r$ are nonnegative integers, by moving upward or to the right along the lattice. Then you must make $n - 1$ rightward moves and $r - 1$ upward moves, for a total of $n - 1 + r - 1 = n + r - 2$ moves. Your final step is either an upward move, in which case $r - 1$ of your first $n + r - 3$ moves have been in the upward direction, or a rightward move, in which case $r$ of your first $n + r - 3$ moves have been in the upward direction. Since your final move is completely determined by whether you are directly to the left or directly below the point $(n, r)$, $$\binom{n + r - 2}{r} = \binom{n + r - 3}{r - 1} + \binom{n + r - 3}{r}$$ However, this argument does not yield the desired formula. Instead, suppose we wish to travel from $(0, 0)$ to $(n - r + 1, r + 1)$ by moving upward or to the right along the lattice. Then you must make $n -r + r = n$ moves, of which $r$ are in the upward direction. If you adapt the argument given above, you should be able to derive the desired formula.

N. F. Taussig
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