In chess960 there are 960 starting positions but i couldnot able to solve theroticaly with the help of permuations.

Both sides(BLACK & WHITE) are same we take only side and we don't consider pawns also.

There are 8 pieces in the board, we can arrange in them in 8! ways.

Since Rooks,Bishops and Knight are repeating then we get 8!/(2!2!2!)=5040.

I am not geting 960 variations?

any MATHEMETICIAN legend help me to solve.

You have to remove the positions where both bishops are on the same colour squares, and where the king isn't between both rooks.

 How to consider these limitaions like King comes always between the rooks

Lemme give it a try.

4 spots for the first Bishop (light sq's)

4 for the second Bishop (dark sq's)

6 spots for the Queen.

5 spots for the first Knight

4 spots for the second Knight

Since these N's are not distinct, we divide by 2

R/K/R occupy the remaining three squares in that order, so we leave 'em out of the probabilities.

(4*4*4*6*5) / 2 = 960

Is this mathematically correct?

It is indeed. Congratulations! 

 Is this the method chess.com uses?

Huh? What are you talking about? This isn't a 'method'. Chess.com have no need to find this number- it is in the name of the game. (Chess960). Simply put, they will have a database of 960 possible positions, and at the start of the game one is randomly picked.

It is a calculation. If it is correct, it is correct everywhere.

This explains the calculation slightly differently. It may be more easily understood.

"The product 4 x 4 x 6 x 10 gives you 960, the total number of starting positions. The first 4 stands for the number of possible squares for the light-square Bishop (b, d, f & h), and the second 4 for the dark-square Bishop (a, c, e & g). The 6 stands for the number of squares available for the Queen, and the 10 for the number of KeRN codes (legal Fischer combinations of Knights, Rooks and the King). See the KeRN table below. 

KeRN Codes:

0 N N R K R   (Knight, Knight, Rook, King, Rook)

1 N R N K R   (Knight, Rook, Knight, King, Rook)

2 N R K N R   (and so on...)

3 N R K R N

4 R N N K R

5 R N K N R

6 R N K R N

7 R K N N R

8 R K N R N

9 R K R N N"

As a side note, the 10 KeRN codes are equivalent to placing the two knights in the five spaces remaining after placing the bishops and the queen.* After the knights are placed, there is only one way to place the king and the rooks, since there are only three spaces left and the king must be in between the rooks.

* five choose two = 5! / (2! * 3!) = 120 / (2 * 6) = 120 / 12 = 10.

 I mean, does chess.com pick the light bishop, then dark, then Queen, etc.?

I doubt it. The memory required to store all 960 possible positions is minimal, so it would be simpler and quicker just to generate one random number and pick from the list. Unless you didn't want to deal with external data access, in which case placing each piece individually would probably be the easiest way to hard code it.

 Ok...

I'm not into chess960, but does the 960 variations include all possible variations including the usual set up? If so may be it should be called chess959 as I don't think the usual set up is included.

The original position IS included, you have a 1/960 chance of getting it.

anyone care to catalogue the 960 different positions here?

Here's a link to a list of all 960 positions as FEN strings:

http://www.chessvariants.com/diffsetup.dir/fischer-random-fen.html

I'm playing a game right now where the only difference is the king's rook is swapped with the king's bishop. It's interesting, because it's like being able to play the standard game, but with a quick king side fianchetto.

 did NOT see that coming lol...

 We should invent machines that can do repetitive tasks like that for us.