There is 9-9 board of which 46 random squares are selected and colored red. Can anyone of you help me to provehere that there exists a 2-2 block in the board of 4 squares of which 3 are colored red.(adsbygoogle = window.adsbygoogle || []).push({});

Well I proved it. I just wanted to see whether there is any alternative approach. Divide the board into 9 groups of 3-3 blocks. One of the blocks must thus contain 6 squares. If the given condition is not valid in the this block then the middle row must be completely unfilled. There should be atleast one adjoining blocks to one of the completely filled rows. The adjacent row of the other block will thus contain no red squares. Thus the maximum number of red squares the other block will contain is 3. Thus 9 red squars in 2 blocks. 37 red squares are to be remained in the 7 remaining blocks. So 2 more blocks must contain 6 red squares because if any block contain 7 or more red squares then the given condition goes valid. Now there are three sets containing 6 red squares. One block containind three red squares can be adjacent block to two of the blocks but not to the third block. So now there is 1 more block with three red squares. So there 22 remaining red squares in the remaining 4 blocks. So two of the next boxes contain 6 red squares and also the next one contains 3 red squares since 2 blocks of three red squares cannot themselves be adjacent to 5 blocks. And thus now one block is remaining and there are 7 red squares in tat block. So we are done.

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# Homework Help: Combinetorics problem

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