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?

Explanation:

Call the quadrants I, II, III, and IV, counterclockwise, and the kinds of flowers A, B, C, and D.

Start by assigning a flower of kind A to quadrant I.

Then, quadrant II can have flower B, C, or D.

If quadrant III has A, then quadrant IV can have either B, C, or D. That is 3 different gardens.

If quadrant III has C, then quadrant IV can have either B or D. That is 2 different gardens.

If quadrant III has D, then quadrant IV can either either B or C. That is 2 different gardens.

Those are all the possible different gardens if A is assigned to quadrant I and B is assigned to quadrant II: 3 + 2 + 2 = 7.

Working yet with A in quadrant I but assigning C to quadrant II, you can make the same reasoning as above to get other 7 gardens.

The same with A in quadrant I but D in quadrant II, other 7 different gardents.

In total, you find 7 × 3 = 21 different gardens when flowers of kind A are assingned to quadrant I.

Starting with a flower of kind B in quadrant I, there will be other 21 different gardens. The same for C and D. Then, in total, there are 4 × 21 = 84 different possible gardens.

congruent segments

step-by-step explanation: