# Planting Flowers

• November 21st 2006, 11:16 AM
MathMage89
Planting Flowers
Four triangular gardening plots form a square as shown. Each plot will contain one kind of flower, and 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?

This is the best picture I could find except it's "square" shaped rather than a "diamond"...-http://www.rogersconnection.com/images/p32.gif

Any help is great. Thanks :)
• November 21st 2006, 12:00 PM
Plato
There needs to be at least one clarification.
If we plant one such garden and the then rotate it 90degrees is the considered a different planting? You see, if it is then planting RLRL, roses and lilies, to rotate that 90deg we would have a different garden but to rotate it 180deg we have the same garden. But that would not be the case with RCRL.
• November 21st 2006, 12:23 PM
MathMage89
Well, the problem doesn't give anymore information...so I'd say that a new garden shouldn't be counted if you rotate RLRL 180 degrees (because technically the same kind of flower is being planted in each plot), but if you rotate it 90 degrees then that is a new garden.
• November 21st 2006, 01:32 PM
Plato
That does not seem right, but….
There are five ways to plant using all different varieties. So with four rotations each that is 20.

We can choose to plant one set of opposite triangles in five ways and plant the other two in six ways. That is thirty ways with four rotations that gives 120 more.

There are ten ways to choose two varieties to plant each set of opposing triangles. Using only two rotations then we have 20 more.
• November 21st 2006, 01:43 PM
MathMage89
So, what would it be if you counted the 180 degree rotations? I suppose that would be the more correct answer?
• November 21st 2006, 01:54 PM
Plato
Quote:

Originally Posted by MathMage89
So, what would it be if you counted the 180 degree rotations? I suppose that would be the more correct answer?

No I do not agree. I think that we can rotate 0, 90, 180, and 270. That gives the four poaitions that I used in each of the first two cases.

With exactly two types of flowers, we use only 0 and 90. That is two positions.
• November 21st 2006, 02:09 PM
ThePerfectHacker
I do not have much time know (and I am lazy). But I think you can solve this problem by considering an adjanency matrix for this graph. The important result that the value in $A^k$ is the number of walks between two vertices in a graph having length $k$.