The streets of a city are arranged like the lines of a chess-board. Let the dimensions be \(m \times n\). Find the number of ways in which a man can travel from the North-West corner to the South-East corner, going the shortest possible distance.

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The streets of a city are arranged like the lines of a chess-board. Let the dimensions be \(m \times n\). Find the number of ways in which a man can travel from the North-West corner to the South-East corner, going the shortest possible distance.

Mathematics
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thts a lot of turns i can take x(
Kinda sounds like problem asked by FFM
who know when i could get lost :/ @Diyadiya will drive me! then no need to worry!

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It is similar to FFM's problem.
then it must be related to this http://en.wikipedia.org/wiki/Motzkin_number
n+1th Mozkin's no
@Ishaan94 i have an answer
o wait not yet!
|dw:1339345895603:dw| seems to be six int this case
yes...six..\({4\choose 2}\)
maybe I should write it \[{2+2\choose 2}\]
|dw:1339346361572:dw|
|dw:1339346637952:dw|
so for 2x3 ...we have 10
\[{5 \choose 2}\]
so adding one segments ... \[ \binom {4 + 1}{2}\]
\[{3+2\choose 2}\]
|dw:1339346850772:dw|
for your 3x2 you will have to go 3 units up and 2 units to the right in total to get to the upper right hand corner UUURR the number of ways to arrange these letters is \({3+2\choose 2}\)
|dw:1339346924974:dw|
2*4+2*10+6*3 = \( \binom{9}{2} = \binom{3 + 3 + 3}{2}\)
there are only 20 ways for a 3x3
\[{m+n\choose n}\]
Yeah, because you take m+n steps and you must decide which n of them will be right, or which m of them will be down (it's equivalent).
yes
isn't there 36 steps in 3x3??
show me a path that requires 36 steps
something wrong with draw app here ... can't copy my own drawing!!
only 6 steps are needed to get to the bottom right
|dw:1339348259654:dw|
there are 6 ways to do a 2x2 not 10
Oh .. i made mistake then ... it must be 3x2
|dw:1339348688431:dw|
yes
shouldn't it be 2*3*3+2*4+2*6 = 38 ??
try and draw 38 minimum distance paths using that graph
I'm getting ( m + n -2)!/(m-1)!(n-1)! ??????
|dw:1339384651187:dw|
Let each square formed be a unit square. Therefore if a man needs to get from NW to south east, He basically needs to cover "n-1" units on the east line. And "m-1" units south. --- For the Shortest possible path when he decides to never turn back. Now in order to do this each street taken by him must be in the South or east direction. Every time he goes east Lets call that turn/ street taken E. Every time he turns south we call that S. So any path followed by him can be written as: SSESESESSSSEEE........... ----- Meaning the man goes first south then south again without turning at next intersection Then turns east at the next turn.......... Here number of S's = m-1 No. of E's = n-1. So the problem has become an equivilant of Finding the permutations of n-1 E's and m-1 S's. = ( m -1 + n -1)! ----------------- (m-1)!(n-1)!
|dw:1339422624581:dw| this is a 1x1 block so m=1 and n=1 using your formula \[\frac{(1-1+1-1)}{(1-1)!(1-1)!}=\frac{0!}{0!0!}=1\] but there are 2 paths to the end
My formula deals with streets. => lines. => m = 2, n=2 in this caase
mine does not
I go by the number of blocks created and the number of steps to go from one corner to the other for a 1x1 block you have to go down 1 and to the right 1
should have never doubted @Zarkon http://math.stackexchange.com/questions/103470/how-can-i-find-the-number-of-the-shortest-paths-between-two-points-on-a-2d-latti

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