How is calculated the total positions in Chess960?

I heard that Chess960 is called like this because 960 means all positions of this type of game. But how was this number calculated?

http://en.wikipedia.org/wiki/Permutation

Wikipedia can explain it better than me:

"Each bishop can take one of four positions, the queen one of six, and the two knights can assume five or four possible positions respectively. This leaves three open squares which the king and rooks must occupy according to setup stipulations, without choice. This means there are 4×4×6×5×4 = 1920 possible starting positions if the two knights were different in some way. However, the two knights are indistinguishable during play (if swapped, there would be no difference), so the number of distinguishable possible positions is half of 1920, or 1920/2 = 960. (Half of the 960 are left-right mirror images of the other half, however Chess960 castling rules preserve left-right asymmetry in play.)"

Though calculating this number is related to permutation, the link you provided was not as useful for my question, although it can be very useful for students of high school math.

Interesting, but what about i start with the knights? Using the same logic, each knight can take 8 and 7 squares, the queen six, the bishops depends where the other pieces stay. How we do it now?

I sat down & did this calculation several months ago & managed to make it tally - this is how I reasoned it out...

L.S.Bishop ...... 4 squares  (4 permutations in total)
D.S.Bishop ...... 4 squares  (16 permutations in total)
Queen ........... 6 squares  (96 permutations in total)
Knight-pair ..... 10 possible pairings on 5 squares (4+3+2+1)

This leaves 3 squares for king & rooks but since the king must always have a rook on either side (otherwise, how do you castle?), there's only 1 permutation of this.

Gives 960 possible arrangements of pieces!

* Because one knight is indistinguishable from the other, if you start with the knights, all possible permutations are 7+6+5+4+3+2+1 = 28  instead of 8 x 7 = 56 !

Edit:  If you start with all permutations for the knights, it seems as if the calculation doesn't work because 960 can't be divided by 28 but you have to take into account that when both knights are on different-coloured squares, there are 9 permutations of bishops, whereas when the knights are on same-coloured squares, there are only 8 permutations of bishops.

Much better to start with the bishops  
 

Thinking well, it should be renamed to Chess959, since 1 of 960 positions is the standard.

Unless it's actually left out (is it?) why not count it?

It is counted. 1 X 960 = 960.

@wafflemaster, i believe that the original idea of Chess960(or Fischer Random) was a attempt to rupture with the classical chess, avoiding all the openings well-known in the standard position. So, it should be Chess 959.

Either the classic position is or is not allowed in 960.  This is the only thing that determines whether it's counted or not.

Chess 1 (normal chess), vs Chess 960 (variations which include normal chess).

Rooks need not be on opposite sides of the King at start.  Indeed one or both could be in the end castling position at game's start.  Some sites have the rule of plunking the K on top of the R to achieve the castling move, others have you lift and replace the K, still others have differing methods.

sftac

sftac, I havn't played a great many games of 960 but I can't remember one in which both rooks were on the same side of the king at the start. Also, the calculation doesn't come out correctly unless you assume that the king is always placed between the rooks.

If what you say were true, it would be called Chess 2880 instaed so I really have to challenge your claim!

Yeah!

In Chess 960 the king is always between the rooks, this allows kingside and queenside castling similarly to regular chess.

Chess960 is not the only option for random chess. It is always possible to allow any possible starting position for the pieces on the back rank. Then it is even possible to have both bishops starting on the same color.

A little combinatorics.

About any possible position with the pieces on backrank, would be also possible the 2 rooks on the same side(especially with king on the corner), making castling impossible. It wouldn't be the best way to play chess IMO.

   Its called Chess 960 because in the year 960bc, the guy who was playing black was a sore loser; and every yr he came back and said,"let's try it this way"; it took him 960 yrs to finally win a game; ya know, like the Red Sox!

I finally figured it out everyone. But I hadta do it my own way to understand it (& myself).

Since the K must be between the two rooks (rules, rules rules)....then there are (56) ways the KRR combination can be setup. Ask me back if u need help w/ this (plz allow one year for a response).

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Now. Here's the trick part that gets people every time (kinda like on Halloween). 

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You must have the bishops of opposite colors thingy going on (more rules). So, let's do those next.

There are five vacant squares for the BB's to fill AFTER the KRR's are in place, right ? So, the combos for BB is six times per KRR setup (accounting for them being opposite color).

BUT (....think big huge BUTT here - lol !), when the KRR all occupy the SAME color square ?....there are only four possible ways the BB combo works - not six. This happens 16 times (ur just gonna hafta trust me on this one (sorry about that)....tho' if uv come this far w/ me ?....then u can probably work it out w/ pencil & pad (....u may wanna lick ur pencil).

So....56 (KRR combos) x 6 (BB combos AFTER the KRR has been set down) MINUS the 16 ways the BB combo CANNOT happen.

(56 x 6) - 16 = 320 ways....so far.

Now. the easy part. There are three squares still vacant for the QNN combo. The combos are: QNN or NQN or NNQ. That's the three distinct ways (as the NN's DO NOT needta be opposite color (go finger).

So....320 x 3 = 960....voila !!

Ahh, things a tiny mind does to amuse itself in bed  .