## Main Activity

Suppose we want to build a brick wall out of the usual size of a brick which has a length of 2 units and a height of 1 unit.

If our wall is to be 2 units tall, we can build our wall in a number of different ways depending on how long we want to build it. Our problem here is to find the number of ways that we can build a 10 units long wall.

Use the canvas here to study the problem:

We may start solving the problem by arranging the bricks for a 2 x 1 rectangle first.

How many possible ways can they be laid?

Next, look at the 2 x 2 construction.

How many arrangements are possible?

Now try 2 x 3 arrangements.

Before we work our way up to the 2 x 10. Let’s try to see if any particular pattern starting to emerge?

## Solution

The tricky pattern to spot is that each total on the right is the sum of the previous two totals.

- 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …

You may have heard of the Fibonacci numbers will notice that the totals are the Fibonacci sequence, starting at 1, 2, …

So by following the Fibonacci Sequence, for the 10-unit long wall; there are 89 possible ways to arrange the bricks

You may read more about Fibonacci Numbers on Mathigon

## Extension

We may wonder in how many different ways can we build an “n” unit-long wall.

In fact, there is a formula for finding any Fibonacci number without having to add up all the previous ones.

If we go back to the first question, for the 10-unit-long wall;

the number with 5 pairs lengthwise and no slabs across the path (just 1 way)= 5 choose 5

the number with 4 pairs lengthwise and 2 across (15 ways)= 6 choose 4

the number with 3 pairs lengthwise and 4 across (35 ways)= 7 choose 3

the number with 2 pairs lengthwise and 6 across (28 ways)= 8 choose 2

the number with one pair lengthwise and 8 across (9 ways) = 9 choose 1

and the number with no pairs lengthwise and all 10 across (1 way) = 10 choose 0

So the total number of ways is

C(10,0) + C(9,1) + C(8,2) + C(7,3) + C(6,4) + C(5,5)

1 + 9 + 28 + 35 + 15 + 1 = 89

We may see these numbers on the Pascal Triangle in a diagonal format.

So for n unit long wall, we can use the combination idea to choose the lengthwise bricks from a total of n bricks.

The total number of ways:

C(n,0) + C(n-1,1)+ C(n-2,2)+ C(n-3,3)+ C(n-4,4) ...+ C(n,r) where r cannot be bigger than n

We can try our latest discovery for a 12 units long wall;

With Fibonacci sequence; 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, **233**

With the Combination method;

**C(12,0)**** + ****C(11,1****) +**** C(10,2)**** + ****C(9,3)**** + ****C(8,4)**** + ****C(7,5)**** + ****C(6,6)**** >>**

**1**** + ****11**** + ****45**** + ****84**** + ****70**** + ****21**** + ****1**** = 233**