Basically assume the three steps to be l (left), d (diagonal), u(up). U cant
have down and right.
Now assume its a 3X3 board.
Then u have (a single d replaces one u and one l)
uuulll, uudll, uddl, ddd.
The total number of permutaions of all the above will give u the result.
The same implies for NXN.
For rook it will be only uuullll hence 6!/(3!*3!)
Whereas for bishop ddd, hence only 1, (as bishop is not allowed to back).
On 5/9/07, Phanisekhar B V <[EMAIL PROTECTED]> wrote:
>
> Sorry that not for queen
>
> On 5/9/07, Phanisekhar B V <[EMAIL PROTECTED]> wrote:
> >
> > Oops, (2n)!/((n!)^2)
> >
> > On 5/9/07, Phanisekhar B V < [EMAIL PROTECTED]> wrote:
> > >
> > > (2n)!/(n!*2)
> > >
> > >
> > > On 5/8/07, PopUp < [EMAIL PROTECTED] > wrote:
> > > >
> > > >
> > > > Hi,
> > > > Consider the chess as a two dimensional array. How will I find the
> > > > number of ways in which queen can reach (n,n) from (0,0). Only
> > > > up,down,diagonal moves are allowed(obviously no back moves).
> > > >
> > > > PopUp
> > > >
> > > >
> > > > > > > >
> > > >
> > >
> >
>
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