In the figure above with 9 squares, how many ways are there to select two squares which do not
share an edge?
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If we want to choose 2 squares from these 9 squares without any restriction ,
It can be done by ( 2 9 ) = 7 ! × 2 ! 9 ! = 2 9 × 8 = 3 6 ways.
But now we will use complementation. We will subtract the ways we don't want the squares to be selected from the total.
Consider the vertical column. If we consider the squares sharing an edge as 1 unit , then there are 4 pairs of squares in vertical column from which 1 can be chosen in ( 1 4 ) = 4 ways. ( A , B ) , ( B , E ) , ( E , H ) , ( H , I )
Similarly , also from horizontal row = ( 1 4 ) = 4 ways. ( C , D ) , ( D , E ) , ( E , F ) , ( F , G )
Hence , total ways = 3 6 − 4 − 4 = 2 8