Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Previous question: link

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?

share|improve this question
    
@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
show 4 more comments

1 Answer

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.