7.1. Draw a row of 11 circles, each of which is either red, blue, or green. Moreover, among any three consecutive circles, there should be at least one red, among any four consecutive circles, there should be at least one blue, and there should be more than half green. How many red circles did you get?
Answer: 3 red circles

Hint. The circles are arranged only as follows: ZZKSKZKSKZZ.

Solution. (1) Three non-overlapping triplets of circles can be identified, each containing at least one red circle. Therefore, there are no fewer than three red circles. (2) Two non-overlapping quartets of circles can be identified, each containing at least one blue circle. Therefore, there are no fewer than two blue circles. (3) According to the condition, there are no fewer than six green circles. (4) Since $3+2+6=11$, all the constraints must turn into equalities. In particular, there are 3 red circles.