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# Constructive Counting and Counting with Restrictions

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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?

Apr 10, 2021

#1
+118626
+1

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

Apr 11, 2021
#2
+321
0

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."

itachiuchihakujo  Apr 11, 2021
#3
0

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.

Apr 20, 2021