## Edited version of a known problem

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Masum
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### Edited version of a known problem

Probably most of you have solved the following problem:
In how many ways can you finish at $(n,m)$ from $(0,0)$ if the possible moves are only up and right. Now the question is what would be the number of ways if you are restricted the area enclosed by the rectangle with vertices $(a,0),(m,0),(m,b),(a,b)$.
One one thing is neutral in the universe, that is $0$.

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### Re: Edited version of a known problem

(Guessing the start point is $(a,0)$ and the end point is $(m,b)$)
Translate $(a,0)$ to $(0,0)$ and solve.
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*Mahi*
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### Re: Edited version of a known problem

Phlembac Adib Hasan wrote:(Guessing the start point is $(a,0)$ and the end point is $(m,b)$)
Translate $(a,0)$ to $(0,0)$ and solve.
Nah bhai :p
I think Masum bhai means you can't step inside the rectangle formed by those 4 vertices.

Use $L^AT_EX$, It makes our work a lot easier!

Masum
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### Re: Edited version of a known problem

yeah, that area is restricted, so you can't step inside the rectangle.
One one thing is neutral in the universe, that is $0$.

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### Re: Edited version of a known problem

Note that if we move the point $(0,0)$ to $(m,n)$ then it will take $m$ right operation and $n$ up operation .
So the number of required ways is number of ways to choose $m$ numbers from $m+n$ distinct numbers . I mean
$\displaystyle \binom {m+n}{m}$.

Now we will count in how many ways one can move the point through the restricted area .
If one move the point through restricted area then he must move the point through any of points
$(a+k,b-k)$. [$a+k \leq m$ and $b-k \geq 0$]

Now one can move the point from $(0,0)$ to $(a+k,b-k)$ in $\displaystyle \binom {a+b}{a+k}$ ways . And as Adib told using axis transformation one can move the point $(a+k,b-k)$ to $(m,n)$ in $\displaystyle \binom {m+n-a-b}{m-a-k}$

So we can move $(0,0)$ to $(m,n)$ through $(a+k,b-k)$ in $\displaystyle \binom {a+b}{a+k} \binom {m+n-a-b}{m-a-k}$ ways .

Case 1: $b>m-a$
then $k_{max} =m-a$ implying the numbers of ways as $\displaystyle \binom {m+n}{m}- \sum_{k=1}^{m-a} \binom {a+b}{a+k} \binom {m+n-a-b}{m-a-k}$

Case 2: $b\leq m-a$
then $k_{max} =b$ implying the numbers of ways as $\displaystyle \binom {m+n}{m}- \sum_{k=1}^{b} \binom {a+b}{a+k} \binom {m+n-a-b}{m-a-k}$
$\frac{1}{0}$