This is the same as above, but with only 24 available squares for the bricks. How many ways can I distinctly put 5 identical bricks on a 5x5 playing board with either 1 or 0 bricks per square for all squares except the top left one where there cannot be any bricks? So there are 25 * 24 * 23 * 22 * 21 ways to place the bricks = 25! / 20!Īs the bricks are identical, there are 5! ways of placing them (5 options for place 1, 4 for place 2 etc.) so there are 25! / 20! ways of placing distict bricks and 25! / (5! * 20!) ways of placing identical bricks on a 25x25 board. The first brick can be put on any of 25 squares, the second on any of 24, the third on any of 23, the fourth on any of 22 and the fifth on any of 21. How many ways can I distinctly put 5 identical bricks on a 5x5 playing board with either 1 or 0 bricks per square?
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