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2007 Alabama ARML TST Problems/Problem 11

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Problem

In how many distinct ways can a rectangular 3\times 17 grid be tiled with 17 non-overlapping 1\cdot 3 rectangular tiles?

Solution

In both solutions we consider the grid to have 3 rows and 17 columns.

Solution 1

There are either 17 vertical tiles, 14 vertical and 3 horizontal, 11 vertical and 6 horizontal, etc. The horizontal tiles always have to form blocks 3\times 3.

If we now consider tilings that consist of a vertical tiles and b blocks of three horizontal tiles, it is clear that there are a+b \choose b such tilings. Hence we get that the total number of tilings in our case is:

\dfrac{17!}{17!}+\dfrac{15!}{14!}+\dfrac{13!}{11!\cdot 2!}+\dfrac{11!}{8!\cdot 3!}+\dfrac{9!}{5!\cdot 4!}+\dfrac{7!}{2!\cdot ...

It isn't that such a pain to compute, so we do:

1+15+78+165+126+21=\boxed{406}

Solution 2

Let T_n be the number of ways to tile a 3\times n rectangle.

Obviously T_0=T_1=T_2=1.

Consider a 3\times n rectangle with n\geq 3. Take a look at the tile that contains the upper right corner. It can be either vertical or horizontal. If it is vertical, we still need to tile the remaining n-1 columns. If it is horizontal, the other two cells in the rightmost column must be covered by horizontal tiles as well. We are then left with n-3 columns to tile.

This gives us the recurrence T_n = T_{n-1} + T_{n-3} for n\geq 3.

Using it we can now compute:

  n | T(n)
------------
  3 |   2
  4 |   3
  5 |   4
  6 |   6
  7 |   9
  8 |  13
  9 |  19
 10 |  28
 11 |  41
 12 |  60
 13 |  88
 14 | 129
 15 | 189
 16 | 277
 17 | 406

See also

2007 Alabama ARML TST (Problems)
Preceded by:
Problem 10
Followed by:
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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