- #1

- 18,633

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"I wonder how many ways (permutations) are there to set up the board to begin a normal game?"

Being Timmy Thomas he never actually got round to working it out.

Can you work out the answer for the lazy swine?

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- Thread starter Greg Bernhardt
- Start date

- #1

- 18,633

- 8,579

"I wonder how many ways (permutations) are there to set up the board to begin a normal game?"

Being Timmy Thomas he never actually got round to working it out.

Can you work out the answer for the lazy swine?

- #2

- 203

- 0

The first one I get to that hasn't been answered already, and I don't know it!!

:D

I will give it a shot, though.

I came up with 645,120 different combinations.

That number feels WAY OFF though.

- #3

- 203

- 0

I am lost, but I gave it a shot.

I can't wait to see how it is actually supposed to be done.

- #4

- 203

- 0

(I think I got it, anyway)

208089907200 possible combinations.

There are 40320 possible combinations of pawn placements on one side.

Switch the rooks, that makes 80640.

Switch the knights, that makes 161280.

Switch the bishops, that makes 322560 different possible combinations on one side.

For each of thiose, there are 322560 different possible combinations on the other side.

322560^2 = 104044953600.

Switcs sides:

104044953600 * 2 = 208089907200 possible combinations.

[?] [?]

- #5

jamesrc

Science Advisor

Gold Member

- 476

- 1

I think you got it right too, but here's an attempt at being pedantic anyway:

For white, the number of setups on one side of the board is P*R*N*B*Q*K = 8!*2!*2!*2!*1!*1! = 322560

where the letters indicate the number of possible permutations for the named piece.

Black has the same number of possibilities, giving the total 322560*322560 = 104044953600

If Timmy is using a regulation chess board, the ranks will be numbered 1 through 8 and the files will be lettered a through h. White has to be on ranks 1 and 2 while black sets up on ranks 7 and 8, so you can't switch sides. So my final answer is 104044953600 permutations.

** EDIT: ** But that was your second answer anyway, so I guess you win either way. D'oh.

For white, the number of setups on one side of the board is P*R*N*B*Q*K = 8!*2!*2!*2!*1!*1! = 322560

where the letters indicate the number of possible permutations for the named piece.

Black has the same number of possibilities, giving the total 322560*322560 = 104044953600

If Timmy is using a regulation chess board, the ranks will be numbered 1 through 8 and the files will be lettered a through h. White has to be on ranks 1 and 2 while black sets up on ranks 7 and 8, so you can't switch sides. So my final answer is 104044953600 permutations.

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- #6

- 203

- 0

(not a math wiz, obviously)

I had a helluva time trying to come up with a simple equation for how many possible combinations there are for 8 pawns on 8 squares.

What I ended up with (in a round about way, through trial and error) was this:

1 pawn= 1 possibility

2.......2*1=1

3.......2*3=6

4.......6*4=24

5.......24*5=120

6.......120*6=720

7.......720*7=5040

8.......5040*8=40320

Is there some simple linear relationship I am missing here?

I know that if you had 8 numbers to go into 8 squares, and repetition is allowed, your possible combinations would be 8^8.

But this doesn't look quite that simple.

Or is it, and I am just missing something?

- #7

NateTG

Science Advisor

Homework Helper

- 2,450

- 6

If you have n identical pieces on m squares then you have m!/(n! (m-n)!) possible combinations. If you have n different pieces on m squares that's m!/(m-n)! possibilities.

The problem, as posed, is woefully unclear about what a "normal game" is. Do the pieces need to start on the back two ranks? Can the game start in check? Do the bishops need to be on different colors?

In addition, the problem does not clarify whether the pawns, or other pieces are numbered.

Without any restrictions, we can calculate the number of starting arrangements of pieces on the board:

Then we would have:

64!/(8!*56!)*

56!/(8!*48!)*

54!/(2!*52!)*

52!/(2!*50!)*

50!/(2!*48!)*

48!/(2!*46!)*

46!/(2!*44!)*

44!/(2!*42!)*

42!/(38!)

Which telescopes to

64!/(2!^{6}*8!^{2}*38!)

Of course, this list does not eliminate the illegal starting positions like starting with a pawn on the 8th rank, and starting in check, neither of which is legal.

Since I don't see an easy way to deal with the problem above, I'll look at the 'starting in the regular area' problem for a moment:

It's quite easy to solve if the pawns are on the second rank. Now we've got eight squares per side, and the number of arrangements is:

8!/(2!*6!)*

6!/(2!*4!)*

4!/(2!*2!)*

2!/0!

all squared

some simple cancellation gives:

(8!/(2!*2!*2!)^{2}=

7!^{2}

or 25401600

If we throw in the pawns on each side naively, that adds a factor of (16!/(8!*8!))^{2} for a total of

4207442279040000 positions.

Unfortunately, there are a large number of positions in that list that have the king in check. I can't think of any good approach to this problem other than splitting the situation into three cases for each side:

The king on the back rank - no check possibility

The king on the outside four files and on the front rank - checks from opposing bishop,queen (2 positions), and rook possible.

rank - checks from opposing bishop and queen possible.

Technically, you could claim that 'normal game of chess' refers to FIDE rules, and that the permutations refers to the number of rearangements that could occur if the pieces were all numbered without upsetting the starting set up. That number is:

8!^{2}*2!^{6}=13005619200

[EDIT: I misread the question. The answer is 13,005,619,200, 26,011,238,400 if you allow rotations of the board]

The problem, as posed, is woefully unclear about what a "normal game" is. Do the pieces need to start on the back two ranks? Can the game start in check? Do the bishops need to be on different colors?

In addition, the problem does not clarify whether the pawns, or other pieces are numbered.

Without any restrictions, we can calculate the number of starting arrangements of pieces on the board:

Then we would have:

64!/(8!*56!)*

56!/(8!*48!)*

54!/(2!*52!)*

52!/(2!*50!)*

50!/(2!*48!)*

48!/(2!*46!)*

46!/(2!*44!)*

44!/(2!*42!)*

42!/(38!)

Which telescopes to

64!/(2!

Of course, this list does not eliminate the illegal starting positions like starting with a pawn on the 8th rank, and starting in check, neither of which is legal.

Since I don't see an easy way to deal with the problem above, I'll look at the 'starting in the regular area' problem for a moment:

It's quite easy to solve if the pawns are on the second rank. Now we've got eight squares per side, and the number of arrangements is:

8!/(2!*6!)*

6!/(2!*4!)*

4!/(2!*2!)*

2!/0!

all squared

some simple cancellation gives:

(8!/(2!*2!*2!)

7!

or 25401600

If we throw in the pawns on each side naively, that adds a factor of (16!/(8!*8!))

4207442279040000 positions.

Unfortunately, there are a large number of positions in that list that have the king in check. I can't think of any good approach to this problem other than splitting the situation into three cases for each side:

The king on the back rank - no check possibility

The king on the outside four files and on the front rank - checks from opposing bishop,queen (2 positions), and rook possible.

rank - checks from opposing bishop and queen possible.

Technically, you could claim that 'normal game of chess' refers to FIDE rules, and that the permutations refers to the number of rearangements that could occur if the pieces were all numbered without upsetting the starting set up. That number is:

8!

[EDIT: I misread the question. The answer is 13,005,619,200, 26,011,238,400 if you allow rotations of the board]

Last edited:

- #8

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one_raven with the point!

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