I'm stuck with this question
We have eight $1\times 2$ tiles that each one of them has one $1\times 1$ blue square and one $1\times 1$ red square. We want to cover a $4\times 4$ area with these tiles in a way that every row and every column of this area had exactly $2$ blue and $2$ red $1\times 1$ squares. In how many ways we can do this?
First of all, I tried to find out all possible ways which I can cover a $4\times 4$ square by $1\times 2$ dominoes. If I know all such possibilities, then for each such a covering, by obtaining the number of ways that some one can tiles this dominoes in blue-red squares in the manner that described above, we can get the answer. In the article How Many Ways Can We Tile a Rectangular Chessboard With Dominoes? the writer claims that the number of ways which we can tile a $4\times 4$ rectangle is $36$. But he did not described all this $36$ ways.
The above method is very prolix and also, I need to obtain all the possibilities of covering $4\times 4$ square by $1\times 2$ tiles and for each of such covering, all the ways which can satisfied by the conditions of question. Dose any one know a simpler method?
I should point out that the above question is designed for high school students and so I think it must have a simple solution.
Best Answer
It actually should not be too bad (if a little tedious) to compute case-by-case with judicious use of symmetry.
I'll use a $4 \times 4$ matrix with entries $b$ (for blue), $r$ (for red) and $\cdot$, to denote the number of tilings of the $\cdot$ spaces given the colours of the $b$ and $r$ spaces. By symmetry, the cases where the top left square is either $b$ or $r$ and either in a horizontal or vertical domino have equal tiling counts, thus
$$ \text{answer} = 4 \pmatrix{b & r & \cdot& \cdot\cr \cdot & \cdot& \cdot& \cdot\cr\cdot &\cdot &\cdot &\cdot\cr\cdot & \cdot& \cdot&\cdot\cr}$$ Now the other two elements in the top row could either be in one domino (which could be in either orientation, or two vertical dominos, of which the second must be the reverse of the first. By symmetry, these cases with vertical dominos have the same counts. Thus $$ \pmatrix{b & r & \cdot& \cdot\cr \cdot & \cdot& \cdot& \cdot\cr\cdot &\cdot &\cdot &\cdot\cr\cdot & \cdot& \cdot&\cdot\cr} = \pmatrix{b & r & b& r\cr \cdot & \cdot& \cdot& \cdot\cr\cdot &\cdot &\cdot &\cdot\cr\cdot & \cdot& \cdot&\cdot\cr} + \pmatrix{b & r & r & b\cr \cdot & \cdot& \cdot& \cdot\cr\cdot &\cdot &\cdot &\cdot\cr\cdot & \cdot& \cdot&\cdot\cr} + 2 \pmatrix{b & r & r & b\cr \cdot & \cdot & b& r\cr\cdot &\cdot &\cdot &\cdot\cr\cdot & \cdot& \cdot&\cdot\cr} $$
Next, consider the first term on the right above. The (2,1) entry could be $b$ in a horizontal domino (in which case there must be two $(r,b)$ horizontal dominos below it), or $r$ in a horizontal domino (related by symmetry to the last term above), or $b$ or $r$ in a vertical domino (in either case with a $(r,b)$ horizontal domino below it). Thus $$ \pmatrix{b & r & b& r\cr \cdot & \cdot& \cdot& \cdot\cr\cdot &\cdot &\cdot &\cdot\cr\cdot & \cdot& \cdot&\cdot\cr} = \pmatrix{b & r & b& r\cr b & r & \cdot& \cdot\cr r &b &\cdot &\cdot\cr r & b& \cdot&\cdot\cr} + \pmatrix{b & r & r & b\cr \cdot & \cdot & b& r\cr\cdot &\cdot &\cdot &\cdot\cr\cdot & \cdot& \cdot&\cdot\cr} + \pmatrix{b & r & b& r\cr b & \cdot& \cdot& \cdot\cr r &\cdot &\cdot &\cdot\cr r & b & \cdot&\cdot\cr} + \pmatrix{b & r & b& r\cr r & \cdot& \cdot& \cdot\cr b &\cdot &\cdot &\cdot\cr r & b & \cdot&\cdot\cr} $$
I'll leave the rest to you...