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(Q). In how many ways two queens can be placed on a 8×8 chess board such that they are not able to attack each other (Queens can attack in the same row/column/diagonal)?
(i). one queen is black & other queen is white
(ii). both queens are identical.
Solution:- we will solve this question with the help of diagram and taking different cases:-

1 queen

case(i): This 8×8 chess board has 4 center squares.
when one  of the queen is placed on one of these squares then number of ways to place 1ˢᵗ queen = 4
Here 4 check (✓) marks & 24 cross (✖) marks show the moves of 1ˢᵗ queen.
Hence 2ⁿᵈ queen can not be placed at these places.
∴ Number of ways to place 2ⁿᵈ queeen = 64 – (4 + 28)
= 36
So total number of ways in this case = 4×36 = 144

2 queen

case (ii): When 1ˢᵗ queen is placed at one of the squares as shown in the above figure by check marks (✓), that can be done in 12 ways.
Here dots (•) & (✖) cross marks shows the moves of the 1ˢᵗ queen.
So 2ⁿᵈ queen can be placed at 64 – 26 = 38 places
So total number of ways in this case = 12×38 = 456

2 queen

case (iii): When 1ˢᵗ queen is placed at one of the squares as shown in the given figure by check marks (✓), that can be done in 20 ways.
Here dot (•) & cross marks (✖) show the moves of the first queen.|
So 2ⁿᵈ queen can not be placed at 12 (•) + 12 (✖) = 24 places
∴  2ⁿᵈ queen can be placed at 64 – 24 = 40 places

So total number of ways in this case = 20×40 = 800

2 queen

case (iv): When 1ˢᵗ queen is placed at one of the given squares as shown in the given figure by check sign (✓) that can be done in 28 ways.
Here dots (•) & cross marks (✖) show the moves of the 1ˢᵗ queen.
So 2ⁿᵈ queen can not be placed at 16 (•) + 6 (✖) = 22 places
∴ 2ⁿᵈ queen can be placed at 64 – 22 = 42 places
So total number of ways in this case = 28×42
                                                                 = 1176

Hence (i). when one queen is black & other is white then desired result can be achieved in 144 + 456 + 800 + 1176
      = 2576 ways           Answer

(ii). When both queens are identical
Then we can not distinguish between queens. Hence in this case the count will become half of the previous count when both queens are distinguishable
Hence desired result =  \frac{{2576}}{2}  = 1288     Answer