# Thread: Possible arrangements

1. ## Possible arrangements

Four congruent triangular gardening plots form a square. Each plot will contain one kind of flower, and the flowers in plots that share an edge will be different. How many different ways can the garden be planted if the flowers can be roses, carnations, daisies, lilies, or tulips? Help would be appreciated thank you!

2. Hello, jarny!

Four congruent triangular gardening plots form a square.
Each plot will contain one kind of flower, and the flowers in plots
that share an edge will be different.
How many different ways can the garden be planted if the flowers can be:
roses, carnations, daisies, lilies, or tulips?

I'll make it easier for me. .I will assume that rotations of the plots
. . are considered different arrangements.

The four plots look like this:
Code:
      * - - - - - - - *
| \           / |
|   \   1   /   |
|     \   /     |
|  4    *    2  |
|     /   \     |
|   /   3   \   |
| /           \ |
* - - - - - - - *

There are five types of flowers; call them: $A,\,B,\,C,\,D,\,E.$

There are $\boxed{5}$ choices for plot #1.
. . Suppose it is flower $A.$
Then we have:
Code:
      * - - - - - *
| \   A   / |
|   \   /   |
| 4   *   2 |
|   /   \   |
| /   3   \ |
* - - - - - *

Then there are $\boxed{4}$ choices for plot #2.
. . Suppose it is flower $B.$
Then we have:
Code:
      * - - - - - *
| \   A   / |
|   \   /   |
| 4   *   B |
|   /   \   |
| /   3   \ |
* - - - - - *

Now there are $4$ choices for plot #3.
But we must consider two possible cases.

[1] Plot #3 has flower $A.$
Then we have:
Code:
      * - - - - - *
| \   A   / |
|   \   /   |
| 4   *   B |
|   /   \   |
| /   A   \ |
* - - - - - *

and there are $4$ choices for plot #4.

Hence, there are: . $5 \times 4 \times 4\:=$ 80 ways.

[2] Plot #3 does not have $A.$
It has $C,\,D,\text{ or }E,\;\boxed{3}$ choices.
Suppose it is flower $C.$
Then we have:
Code:
      * - - - - - *
| \   A   / |
|   \   /   |
| 4   *   B |
|   /   \   |
| /   C   \ |
* - - - - - *

Now there are $\boxed{3}$ choices for plot #4.

Hence, there are: . $5 \times 4 \times 3 \times 3 \:=$ 180 ways.

Therefore, there are: . $80 + 180 \:=$ 260 ways.