Rook-Chessboard problem. 6 rooks on 8 by 8 board

[m1sanch]

Homework Statement

How many ways can two red and four blue rooks be placed on an 8-by-8 chessboard so that no two rooks can attack one another.

Homework Equations

1)Number of permutations with two types=(n choose n₁)=n!/((n₁!)(n-n₁1))
2)Number of ways to place n rooks which have k object types(colors) on n-by-n board=(n!)²/((n₁!)(n₂!)...(n_k!))

The Attempt at a Solution

With the two equations on top I thought that they answer would be (8 choose 6)((8 choose 6) choose 2)) since it will be a 6 rooks on an 8-by-8 board.

*(8 choose 6)=(8!)/(6!((8-6)!)=8x7/2=28
This will simply to:
28(28!/(2!(28-2)!)=
28(28x27/2)=
14x28x27=28x27x14.

Did I do this correct or am I wrong in my reasoning.


P.S. I am sorry for the format. I am new to this site and tried to make it as clean as possible.

 

[willem2]

I don't see what you're doing here.
The answer should be a lot bigger however.

It's possible to choose the both the columns and the rows that will have blue and red rooks independently.
The number of ways to choose the colums is the number of ways of ordering (b,b,b,b,r,r,e,e) (e=empty).
Same for the rows.
No matter which columns and rows you choose you will end up with a 4x4 square of possible locations for the blue rooks and a 2x2 square of possible locations for the red rooks.

 

[m1sanch]

Ok so I can regard my multiset of "rooks" as being a set m={4*b,2*r,2*e} like you stated and use this theorem in my book that states:

For n rooks of k colors the number or arrangements is equal to

n!(n!/(the repetition number of each object)!).

So the number of arrangements would be (8!)²/(4!*2!*2!).

Which equals 10*7*6*8! Since 8!/(4!*4)=10*7*6.

 

[kduna]

You could just count directly.

With 6 non-attacking rooks on an 8x8 board, there will be exactly two rows and two columns that are free of rooks. Choose the two rows, there are (8 choose 2) ways to do so. Now choose the rows for your 4 blue rooks, there are (6 choose 4) ways to do so. At this point the rows of the remaining two rooks has already been determined by your choices. Now think of your six rooks and two empty places for a total of 8 objects. Take any of the 8! permutations of these and throw them in columns that way to get a non-attacking position. Thus the number of possible positions is:

((8 choose 2)*(6 choose 4))*8!​

 

[m1sanch]

Oh ok thank you for the other alternative to look at this problem. It does make sense now.

 

[haruspex]

You could just count directly.

With 6 non-attacking rooks on an 8x8 board, there will be exactly two rows and two columns that are free of rooks. Choose the two rows, there are (8 choose 2) ways to do so. Now choose the rows for your 4 blue rooks, there are (6 choose 4) ways to do so. At this point the rows of the remaining two rooks has already been determined by your choices. Now think of your six rooks and two empty places for a total of 8 objects. Take any of the 8! permutations of these and throw them in columns that way to get a non-attacking position. Thus the number of possible positions is:

((8 choose 2)*(6 choose 4))*8!​

A factor of 2 too many. When choosing the columns, the two blank rows are interchangeable.

 

[willem2]

Ok so I can regard my multiset of "rooks" as being a set m={4*b,2*r,2*e} like you stated and use this theorem in my book that states:

For n rooks of k colors the number or arrangements is equal to

n!(n!/(the repetition number of each object)!).

So the number of arrangements would be (8!)²/(4!*2!*2!).

Which equals 10*7*6*8! Since 8!/(4!*4)=10*7*6.

8! <> 10 so the arithmetic here is wrong. (8!)²/(4!*2!*2!) is a factor of 2 too high.

You'd have 8!/(4!2!2!) arrangements for the rows, and the same number for the columns, and than 4! ways to order the blue and 2! ways to order the red rooks, which ends up being equal to
(8!/4! 2! 2!)^2 * 4! * 2! which is equal to ((8 choose 2) (6 choose 4) * 8!) / 2 as has been given by kduna and haruspex.