+321 A gardener has a circular garden, which he divides into four quadrants. He has four different kinds of flowers, and he wants to assign a flower to each quadrant so that adjacent quadrants are not planted with the same kind of flower. (Also, the gardener can use the same flower more than once.) How many different gardens are possible?
+129852 Here's my take
Start with the top left quadrant - we have 4 choices
In the top right quad, we have 3 choices (we can't plant the same type as in the top left quad)
In the bottom right quad, we again have 3 choices ( we can choose to plant the type of flower planted in the top left quad as well as the other 2 types ....we can't plant the same type as in the top right quad)
The sane number of choices apply to the bottom left quad as with the bottom right quad
So
4 * 3 * 3 * 3 = 108 possible gardens
+321 it is wrong it came with a hint though "
Find the number of ways to plant quadrants A and C.
Then the number of ways of planting quadrants B and D depends on whether A and C have the same flower or not, so divide into these cases."
see if that works. :0
For this question, we cannot see it as a normal finding problem because this have 2 adjacent side, so here is how you really should do it:
3x2x2x2 :) because A is effected by D so 3 choices only,
B is effected by A and C so 2
C is effected by B and D so 2
, and same for D.
