The streets of a city are arranged like the like the lines of a chess board. There are `m` streets running from north to south and `n` streets from east to west. Find the number of ways in which a man can travel from north-west to south-east corner, covering shortest possible distance.

Explanation

Let the city be represented by a rectangle whose sides are a and b.
As shown in the figure, length a is divided by (m-1) segments and length b is divided by (n-1) segments formed by parallel streets.

To reach from one corner P to the other corner Q in shortest possible path, one has to cover (m-1) segments from North to South and (n-1) segments from West to East. The number of shortest paths is equal to number of ways in which the order of (m-1) N-S and (n-1) W-E segments is decided.
So, required number of shortest paths `=((m+n-2)!)/((m-1)!(n-1)!)`