A 6x6 grid is cut from an 8x8 chessboard. In how many ways can we put two identical coins, one on the black square and one on a white square on the grid, such that they are not placed in the same row or in the same column?

A.

216

B.

324

C.

144

D.

108

• plz explain it ..
• 9 years agoHelpfull: Yes(0) No(1)
• Option(A) is correct
  
  In a 6×6 grid of a chessboard, each row and each column contains 3 white and 3 black squares placed alternatively.
  
  There are a total of 18 black and 18 white squares.
  
  For every black square chosen to put one coin, we cannot choose any white square present in its row or column.
  
  There are 3 white squares in its row and 3 white square in its column for every black square.
  
  Hence for every black square chosen, we can choose (18−6)=12 white squares.
  
  Total number of possibilities where a black square and a white square can be chosen so that they do not fall in the same row or in the same column:
  
  =18×12=216
  So there are 216 ways of placing the coins that are identical.
  
• 8 years agoHelpfull: Yes(0) No(0)