<PRE><B>Question 1160021</B>
&lt;br&gt;
When I read this problem, I thought it was going to be solved by a simple recursive pattern, as in the problem of how many ways can you tile an n-by-2 rectangle with 1-by-2 tiles, or how many ways can you climb a flight of 8 stairs if you can climb either 1 or 2 steps at a time.&lt;br&gt;
But my attempt to find a recursive pattern quickly got ugly.&lt;br&gt;
I turns out this is a MUCH more difficult problem to analyze.  (Although there might be solution methods I don&#39;t see that might be easier than mine....)&lt;br&gt;
So I tried a different approach.&lt;br&gt;
We can break down the analysis in terms of the numbers of 2-by-2 tiles used.&lt;br&gt;
Consider a diagram of the 3-by-8 road and the possible locations of the CENTERS of the 2-by-2 tiles.  Identify each such location with a (row,column) designation.&lt;br&gt;Here is a text-formatted display of the diagram I used to do my analysis:&lt;br&gt;&lt;pre&gt;

    +------+------+------+
    |      |      |      |
    |      |      |      |
    |      |      |      |
    +-----1,1----1,2-----+
    |      |      |      |
    |      |      |      |
    |      |      |      |
    +-----2,1----2,2-----+
    |      |      |      |
    |      |      |      |
    |      |      |      |
    +-----3,1----3,2-----+
    |      |      |      |
    |      |      |      |
    |      |      |      |
    +-----4,1----4,2-----+
    |      |      |      |
    |      |      |      |
    |      |      |      |
    +-----5,1----5,2-----+
    |      |      |      |
    |      |      |      |
    |      |      |      |
    +-----6,1----6,2-----+
    |      |      |      |
    |      |      |      |
    |      |      |      |
    +-----7,1----7,2-----+
    |      |      |      |
    |      |      |      |
    |      |      |      |
    +------+------+------+

&lt;/pre&gt;
The diagram shows that the center of a single 2-by-2 tile can be on any of 7 levels, and in either of 2 columns.&lt;br&gt;
And when there are more than a single 2-by-2 tile, the levels must differ by at least 2.&lt;br&gt;
Finally, note that we don&#39;t need to consider the 1-by-1 tiles; whatever regions are not covered by the 2-by-2 tiles will be covered in only one way by the 1-by-1 tiles.&lt;br&gt;
So now on to the counting....&lt;br&gt;
(1) No 2-by-2 tiles....&lt;br&gt;
That one is easy.  The entire array is covered by 1-by-1 tiles; there is only 1 way to do this.&lt;br&gt;
(2) One 2-by-2 tile....&lt;br&gt;
This one is also rather easy.  The center of the single 2-by-2 tile can be on any of the 7 levels and in either of 2 columns.  7*2 = 14 ways.&lt;br&gt;
(3) Two 2-by-2 tiles....&lt;br&gt;
By enumeration, we find the number of combinations of 2 of the 7 levels, knowing that they cannot be adjacent:&lt;br&gt;
1-3, 1-4, 1-5, 1-6, 1-7; 2-4, 2-5, 2-6, 2-7; 3-5, 3-6, 3-7; 4-6, 4-7; and 5-7.&lt;br&gt;
That&#39;s 15 combinations; and in each of the levels chosen the two 2-by-2 tiles can each be in either of 2 columns.  15*2*2 = 60 ways.&lt;br&gt;
(4) Three 2-by-2 tiles....&lt;br&gt;
Again we enumerate the number of combinations of 3 of the 7 levels, again with the same restriction:&lt;br&gt;
1-3-5, 1-3-6, 1-3-7, 1-4-6, 1-4-7 1-5-7, 2-4-6, 2-4-7, 2-5-7, and  3-5-7.&lt;br&gt;
That&#39;s 10 combinations; and in each of the levels chosen the three 2-by-2 tiles can each be in either of 2 columns.  10*2*2*2 = 80 ways.&lt;br&gt;
(5) Four 2-by-2 tiles....&lt;br&gt;
There is only one way to choose 4 of the 7 levels with no two adjacent: 1-3-5-7.  And on those four  levels each of the 2-by-2 tiles can be in either of 2 columns.  1*2*2*2*2 = 16 ways.&lt;br&gt;
Final total: 1+14+60+80+16 = 171 ways.&lt;br&gt;
ANSWER: There are 171 ways to cover a 3x8 array with 2x2 and 1x1 tiles.&lt;br&gt;
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