Mr. Markus tiles problem

The answer is 28.

If you try it from \­( 3 x 1 m^{2} \­) tiles, you'll find the pattern. Here it is.

\­( 3 x 1 m^{2} = 1\­)

\­( 3 x 2 m^{2} = 1\­)

\­( 3 x 3 m^{2} = 2\­)

\­( 3 x 4 m^{2} = 3\­)

\­( 3 x 5 m^{2} = 4\­)

\­( 3 x 6 m^{2} = 6\­)

\­( 3 x 7 m^{2} = 9\­)

\­( 3 x 8 m^{2} = 13\­)

\­( 3 x 9 m^{2} = 19\­)

\­( 3 x 10 m^{2} = 28\­)

The pattern is $1,1,2,3,4,6,9,13,19,28,...$

using $n = (n-1) + (n-3)$ for $n \ge 4$

So, the answer is $\boxed{28}$

First, note that this is a recurrence problem. Let $a_n$ denotes the number of ways to place $3 \times 1$ tiles on a $3 \times n$ floor. Then, realize that for $n \geq 3$ , you have $2$ ways to place the first few tiles:

1. Simply place 1 tile vertically. (Then there will be $a_{n-1}$ ways to place the rest of the tiles needed. (Since we have left with a $(n-1) \times 3$ block)

2. Second, place 3 tiles horizontally, and you will be left with $a_{n-3}$ ways to arrange. (Since we have left with a $(n-3) \times 3$ block)

Notice that there isn't any other way to tile the first few tiles (e.g. 2 tiles horizontally, because this arrangement forces the next tile placed to be also horizontal, and this becomes Case 2 above).

Therefore, we can formulate a recurrence:

$a_n = a_{n-1} + a_{n-3}$ for all $n\geq 4$ .

We can easily come up with $a_1 = 1, a_2 = 1, a_3 = 2$ . Repeating the steps, we get $a_10 = \boxed{28}$ .

//This is a recurrence //you can make a pattern at the paper..

• The 3 x 1 we can only put 3 x 1 Tiles in one way so 3 x 1 = 1
• The 3 x 2 we can only put 3 x 1 Tiles in one way too 3 x 2 = 1
• The 3 x 3 we can only put 3 x 1 Tiles in two way.( verticaly and horizontaly) 3 x 3 = 2
• The 3 x 4 we can only put 3 x 1 Tiles in three way, 3 x 4 = 3
• The 3 x 5 we can only put 3 x 1 Tiles in four way, 3 x 5 = 4 //Stop in here because we found a pattern.

//are you see the patern?

• in [3 x 4] is [3 x 3] + [ 3 x 1] = 2 + 1 = 3;
• in [3 x 5] is [3 x 4] + [3 x 2] = 3 + 1 = 4;

//so [3 x n] = [3 x n -1] + [3 x n - 3], to finding 3 x 10 (we must using bottom up)

• [3 x 6] = [3 x 5] + [3 x 3] = 4 + 2 = 6;
• [3 x 7] = [3 x 6] + [3 x 4] = 6 + 3 = 9;
• [3 x 8] = [3 x 7] + [3 x 5] = 9 + 4 = 13;
• [3 x 9] = [3 x 8] + [3 x 6] = 13 + 6 = 19;
• [3 x 10] = [3 x 9] + [3 x 7] = 19 + 9 = 28;