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Consider a binary grid of size $4\times 4$, each of cell can either have $0$ or $1$. Among all possible $2^{16}$ arrangement how many arrangement of such grid exist in which each row and column contains even number of $1$s.

Solution which I thought

There will be $2$ possibilities for the answer of this question $1$st all ones in the $4\times 4$ grid that will count up to $1$ possible arrangement and $2$nd possibility will be $2$ ones in each row and column so how can i find the possible arrangement which will have $2$ ones in each row and column.

Am I right?

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    Why is it not $p$ossible that some rows contain four 1s and some rows contain two 1s?2012-07-06

2 Answers 2

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Let the positions be $a_{i,j}$, where $1\le i,j\le 4$. You can fill the $3\times 3$ square in the upper lefthand corner, i.e., positions $a_{i,j}$ with $1\le i,j\le 3$, any way you like. Once those $9$ positions are filled, there is exactly one way to fill the remaining $7$ positions to get an even number of $1$’s in each row and column. Can you see why? (HINT: Fill $a_{4,4}$ last.)

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Hint: how many choices are there for the first column, ignoring the question of rows? You can put each of these in the first three columns, any way you want. Then each row forces the element in the fourth column. You then have to prove that the fourth column will have an even number of ones.