Find the number of ways in which 2 identical kings can be placed on an n x m chessboard so that the kings are not in adjacent squares.

pankaj kumar raman
51 Points
12 years ago

first king can be placed in n*m places

1)one king at corner ,other king will be at = 4  *  [(n*m)-4] places    = 4mn-16

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4 corners   [m*n places - 4 places around]

2)one king at edge, other will be at= [2(n-2)+2(m-2)]  *  (n*m-5) = 2mn²-10n+2m²n-10m-8mn+40

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number of box at edge      m*n places except 5 places around

3)one king at free from corners and edges, other will be at

= [(n-2)*(m-2)]     *        (n*m-8)      =   n²m²-2n²m-2nm²-4nm+16n+16m-32

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number of boxes away from ends           n*m places except 8 places around

sum of all probablities will be= (nm-4)²+6(m+n-4)

kushal satya
37 Points
12 years ago
 A B D C

 LET THE ABOVE BE THE CHESS BOARD. WHEN ONE OF THE KING IS PLACED AT A,B,C,D THEN NO OF POSITIONS AT WHICH THE OTHER KING COULD BE PLACED IS mn-4 (three adjacent and 1 the position of the first king). hence no of ways in this case = 4(mn-4) no of ways when first king is placed between a and b or b and cand so on in the same row or column then no of ways=2(m+n-4)(mn-6) in the last case when the king is placed elsewhere on the board, no of ways =2(mn-m-n)(mn-9) add all the ways to get the net no of ways.

APURV GOEL
39 Points
12 years ago

sorry this isnt the ans.

pankaj kumar raman
51 Points
12 years ago

then tell me the right solution

as it was disapproved