We have a $2 \times 3$ rectangle, that we wish to cover using $1 \times 2$ rectangles and $1 \times 1$ squares. How many possible ways are there to do so?

Using $0$ and $3$ rectangles, it appears that there are only $2$ possible combinations. Using $1$ rectangle, we can get $7$ solutions. Determining the number of solutions when two rectangles are used appears to be the tricky part.


With $0$ rectangles there is one way, as you say. However, with $3$ rectangles, there are $3$ ways, not $1$. With $1$ rectangle, there are $7$ ways, as you say.

For $2$ rectangles, they can either both extend in the shorter direction, in $3$ ways, or both in the longer direction, in $4$ ways, or one in each direction, in $4$ ways.

The total is $1+3+7+3+4+4=22$.

[Edit in response to comment:]

Let's say the $2\times3$ rectangle has $2$ rows and $3$ columns. For $2$ rectangles, if they both extend along the columns, they occupy one column each, leaving one column for the squares, and there are $3$ choices for that column. If they both extend along the rows, they have to be in different rows, and each has two possible positions in its row, which makes $2\cdot2=4$ options. If they extend in different directions, one occupies a column, and this can't be the central column, so there are $2$ choices for the column, and in each case there are $2$ choices which row to put the other one in, for another $2\cdot2=4$ options.

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  • $\begingroup$ (+1) I see why there are $3$ ways with $3$ rectangles, but the reasoning with $2$ rectangles still seems a little shaky to me. I guess what I am really asking for is a way to prove it mathematically. $\endgroup$ – George V. Williams Mar 15 '13 at 0:04
  • $\begingroup$ Couldn't you make an argument that there are 2 ways with 3 rectangles since you can turn the original rectangle upside down? $\endgroup$ – Darren Mar 15 '13 at 0:15
  • $\begingroup$ @George: I provided a bit more detail; is that better? $\endgroup$ – joriki Mar 15 '13 at 0:18
  • $\begingroup$ @joriki: Now I feel that the solution is a lot more logical. Thanks! $\endgroup$ – George V. Williams Mar 15 '13 at 0:19
  • $\begingroup$ @Darren: Yes, you could certainly make all sorts of arguments based on the imprecise wording of the question; it specifies neither which ways are equivalent nor whether you're allowed to turn the pieces or the space to be filled. My interpretation was that the pieces can be turned but the space can't, but that may not be what was intended. $\endgroup$ – joriki Mar 15 '13 at 0:21

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