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.

- anonymous

- jamiebookeater

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- karatechopper

thts a lot of turns i can take x(

- experimentX

Kinda sounds like problem asked by FFM

- karatechopper

who know when i could get lost :/ @Diyadiya will drive me! then no need to worry!

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## More answers

- anonymous

It is similar to FFM's problem.

- experimentX

then it must be related to this
http://en.wikipedia.org/wiki/Motzkin_number

- experimentX

n+1th Mozkin's no

- karatechopper

@Ishaan94 i have an answer

- karatechopper

o wait not yet!

- experimentX

|dw:1339345895603:dw|
seems to be six int this case

- Zarkon

yes...six..\({4\choose 2}\)

- Zarkon

maybe I should write it \[{2+2\choose 2}\]

- experimentX

|dw:1339346361572:dw|

- experimentX

|dw:1339346637952:dw|

- experimentX

so for 2x3 ...we have 10

- Zarkon

\[{5 \choose 2}\]

- experimentX

so adding one segments ...
\[ \binom {4 + 1}{2}\]

- Zarkon

\[{3+2\choose 2}\]

- experimentX

|dw:1339346850772:dw|

- Zarkon

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 \({3+2\choose 2}\)

- experimentX

|dw:1339346924974:dw|

- experimentX

2*4+2*10+6*3 = \( \binom{9}{2} = \binom{3 + 3 + 3}{2}\)

- Zarkon

there are only 20 ways for a 3x3

- Zarkon

\[{m+n\choose n}\]

- anonymous

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).

- Zarkon

yes

- experimentX

isn't there 36 steps in 3x3??

- Zarkon

show me a path that requires 36 steps

- experimentX

something wrong with draw app here ... can't copy my own drawing!!

- Zarkon

only 6 steps are needed to get to the bottom right

- experimentX

|dw:1339348259654:dw|

- Zarkon

there are 6 ways to do a 2x2 not 10

- experimentX

Oh .. i made mistake then ... it must be 3x2

- experimentX

|dw:1339348688431:dw|

- Zarkon

yes

- experimentX

shouldn't it be 2*3*3+2*4+2*6 = 38 ??

- Zarkon

try and draw 38 minimum distance paths using that graph

- anonymous

I'm getting ( m + n -2)!/(m-1)!(n-1)! ??????

- anonymous

|dw:1339384651187:dw|

- anonymous

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)!

- Zarkon

|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

- anonymous

My formula deals with streets. => lines.
=> m = 2, n=2 in this caase

- Zarkon

mine does not

- Zarkon

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

- experimentX

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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