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Say there are two points $P_1(a_1,b_1)$ and $P_2(a_2,b_2)$, the number of ways of reaching $P_1$ from the origin is $w_1$ and $P_2$ from $P_1$ is $w_2$. (Here $a_1<a_2$ and $b_1<b_2$.) So the number of ways (say $W$) of reaching $P_2$ from the origin through $P_1$ is $W=w_1\cdot w_2$. The number of combinations is given by

$$w_1=\binom{a_1+b_1}{a_1},\quad w_2=\binom{(a_2-a_1)+(b_2-b_1)}{a_2-a_1}.$$

You get the above formula for $w_1$ by shifting to make $P_1$ the origin; the shift involves subtracting the coordinates of $P_1$ out of everything.

However: if the number of ways $W$ is given, how do we find a point $P_2$ such its distance from the origin is maximum?

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@anon Is that clear? – Surya Jan 29 '12 at 7:36
The number of ways of reaching p2 through p1 is w1*w2; If there are w1 ways to reach point p1 and for each way, we have w2 ways to reach point p2. Thus w1*w2 – Surya Jan 29 '12 at 7:37
Don't you mean it's $w_1$ and not $p_1$ that is equal to $\displaystyle\frac{(a_1+b_1)!}{a_1! b_1!} = \ ^{a_1+b_1}C_{a_1} = \binom{a_1+b_1}{a_1}$, and similarly for $w_2$ instead of $p_2$? Apart from that, I think the question is clear as stated. – Rahul Jan 29 '12 at 7:44
Not really, you still have a number of serious readability issues, but I think I can at least fix it now. (Also, sorry, I misread the definition $w_2$, you are correct about counting the ways.) – anon Jan 29 '12 at 7:49
Alright, I've fixed it up and subsequently removed my downvote (which I did state was for unintelligibility; that comment is now removed). I did not vote to close. The combinatorics of two points seems irrelevant to your actual question though. – anon Jan 29 '12 at 8:01

Refer the answer given to Reverting the binomial coefficient. You may need to know Stirling's approximation, a related question for that can be found here and use of calculus or some clever trick later on.

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