A probability problem by Josh Rowley
Consider a standard chessboard. Define as the maximum number of counters we can place on such a chessboard so that no 2 adjacent squares both have counters. Also, define as the maximum number of counters we can place on such a chessboard so that no 2 squares which share a vertex both have counters. What is ?
The answer is 48.
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1 solution
For no adjacent squares to be of same counter(lets take them as colors to be simple) we will have half of the chess board as we have a color(black or white) in the chess board. So A=64/2 = 32 Now as no adjacent vertices can touch each other therefore we can take only 4 in a row and only in the 4 columns(as we can't take columns next to each other). Therefore, B=16. Hence answer is 48