1. ## Tilings

Isabel was making rectangular strips 2 units wide using 2 x 1 tiles. She was interested in the number of different strips of each length she could make.
The attached image shows some of her examples arranged according to the units of length. All strips are different even though some are just reflections

Isabel noticed that the number of ways of making strips followed a pattern: The number of strips n units long is alwasy the sum of the numbers of strips of the two previous lengths. eg: Wn is the nuber of strips of length n, W4=W3+W2 Ingeneral her guess is that : W1=1, W2=2 and Wn=W(n-1) +W(n-2) when n is greater than or equal to 3

a)Explain why isabels guess is true
b) isabel experimented with 3x1 tiles, making strips 3 units wide. Finda formula, similar to that above for the number of strips of length n
c) Find a similar formula for the number of ways of making strips of width m and lenght n from mx1 tiles

2. im sorry i forgot to attach the image

well here it is..

im very desperate.

any help would be very very nice thank you

4. OK. Here is a solution to part a), which is explains why Isabel's guess is right.

Isabel calls $W(n)$ "the number of ways to arrange $n$ of the 2x1 tiles into a 2x $n$ strip".

What we have to prove, is:

$W(n) = W(n-1) + W(n-2)$

This looks pretty hard - how are we going to count all the different ways we can arrange the $n$ tiles? Actually, we don't have to - we can just look at the very first couple of tiles!

The arrangements of $n$ tiles can be split up into two groups. (See the figure, below.)

Some will have the first tile on the left is standing upright (call this "A" group). The others will have the first two tiles laying horizontally (call this "B" group).

Now count the arrangements:

• How many "A" arrangements are there? It's exactly equal to how many ways we can arrange the other (n-1) tiles. So there are $W(n-1)$ "A" arrangements.
• How many "B" arrangements are there? This time, it's equal to how many ways we can arrange the other (n-2) tiles. So, there are $W(n-2)$ "A" arrangements.
The full number of arrangments, $W(n)$ is the numbers of "A" and "B" arrangements added together.

So $W(n) = W(n-1) + W(n-2)$. (QED!)

- - - -

You can use a similar argument to do the other bits of the question. See if you can guess how.

5. thanks for helping me with part a!!

but could anyone explain to me how to do b)... im definitely sure that once i have b cleared out, i can do c) by myself as it would be in a similar sort of format.

but i need some help to get started.

Thank you to anyone who can help.

6. For b), the situation is quite similar. Either the first 3x1 block will be upright ("A") or the first three blocks will be laying horizontally in a block.

You can use the same reasoning as above to show:

$W(n) = W(n-1) + W(n-3)$