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Given a 4x4 grid of squares, can you black out exactly six of them so that the number of blacked out squares in each row, as well as in each column, is even.

All solutions are permutations of a single one (i.e. all solutions are obtained from any given one by simply flipping around rows and columns).

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There are two blacked out squares in each of the 1st, 2nd and 3rd rows and columns and no blacked out squares in the 4th row and column.

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  • $\begingroup$ @JohnRobertson I'm not sure I understand the comment "All solutions are permutations of a single one", as you could arrive at a solution by using $(1,1), (1,2), (2,1), (2,4), (4,2), (4,4)$ as well. That seems to me not to be a permutation of the solution given here. $\endgroup$ – GentlePurpleRain Oct 16 '15 at 16:47
  • $\begingroup$ @GentlePurpleRain Yes, every solution can be obtained from one solution by permuting rows and columns. Your example is obtained from the displayed one by switching the 3rd row with the 4th row and the 3rd column with the 4th column. That changes all your 4s into 3s which becomes exactly the solution shown in the answer. $\endgroup$ – John Robertson Oct 16 '15 at 18:04
  • $\begingroup$ @JohnRobertson I see. But isn't that self-evident, since by permuting rows and columns you can basically move the black squares wherever you want? It seems like the statement about permutations is trying to say that all solutions are identical because all solutions have 2 black squares in each row and column, which is the definition of the question. It's like me asking for a 6-digit number containing the digits 1 to 6, and then stating that "all solutions are permutations of a single one." It doesn't really add any information... $\endgroup$ – GentlePurpleRain Oct 16 '15 at 18:18
  • $\begingroup$ Not quite. It isn't hard to prove per se, but it isn't as easy as what you are saying because not every permutation of a 4x4 block can be obtained by permuting rows and columns and just because you get two in the right place in one row using permutations doesn't a priori mean you can get all the others in the right place with permutations without messing up the ones you already got right. I think it is kind of funny that you have made two objections - the first being that I was saying something false, the second being that I was saying something completely self evident. $\endgroup$ – John Robertson Oct 16 '15 at 18:47
  • $\begingroup$ @JohnRobertson I understand now. My problem was that with my first question I misunderstood your statement and assumed it to be false. Once it was explained, I then assumed it to be self-evident. You have now shown me that $\endgroup$ – GentlePurpleRain Oct 18 '15 at 3:38
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If the question meant to ask:

"..can you black out exactly six of them so that the number of blacked out squares in each row, as well as in each column, is equal?",

then the following solution would work:

enter image description here

With 1.5 blackout out in each row and column.

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