1

Find the number of ways to arrange $8$ rooks on a chessboard such that no two of them attack other?

I was thinking it would be $64 \times 49 \times 36 \times 25 \times 16 \times 9\times 4 \times 1$, because after selecting 1 rook, the only squares left for selecting the next rook would be $15, 13, 11, 9 , 7 , 5$ and $3$ less than when selecting for the current rook?

Can someone explain?

Roby5
  • 4,287
Aden
  • 107
  • 1
  • 7
  • 2
    The answer should be $8!$. Try to figure out a bijection that proves this. But your approach also does work once it's debugged--what do you overcount? – cubesteak Jul 01 '16 at 14:40
  • You've counted permutations rather than combinations. So you need to divide your answer by the number of ways to arrange 8 rooks in a line. Additionally, you've missed out $\times 9$ which I'll correct. – it's a hire car baby Jul 01 '16 at 17:03

2 Answers2

6

Obviously you have to place a rook on the each of the $8$ columns. Now for the first column we have $8$ options. For the second we have $7$, as one of the rows is already taken by the previous rook and so on. Eventually you will get that the number of ways to place them is $8! = 40320$.

To see why your reasoning is false note that the rooks aren't distinguishable. Using your idea you count the way of placing $8$ distinguishable rooks on the chessboard. Hence because the rooks can permutate in $8!$ ways between themselves, you need to divide by $8!$ and you will get exactly the same answer.

Anyway if you want to read more about this problem you may want to look here.

Stefan4024
  • 35,843
1

By your method, there are $64$ ways to place the first rook. This attacks $15$ squares, leaving $49$ ways to place the 2nd rook. This attacks 13 squares not attacked by the first rook, leaving $36$ was to place the third rook, and so on...

$64 \times 49 \times 36 \times 25 \times 16 \times 9\times 4 \times 1 = (8!)^2$

But since every rook is identical and it doesn't matter what order we place them, we must divide by the number of orders in which they can be placed, which is $8$ choices for the first rook, $7$ choices for the 2nd rook, $6$ choices for the 3rd rook and so on, so we will divide by $8!$ to convert the permutations into unique combinations:

$$\frac{(8!)^2}{8!}=8!$$

it's a hire car baby
  • 9,243