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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.
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
↑ ↑
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
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
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)
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.
sorry this isnt the ans.
then tell me the right solution
as it was disapproved
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