Let n be a positive integer and suppose 2n dots are placed a

Let n be a positive integer, and suppose 2n dots are placed around a circle, n of which are colored red and n colored blue. Traveling around the circle clockwise, you keep a count of how many red dots and blue dots you have passed. If at all times the number of red dots is greater than or equal to the number of blue dots, you consider your trip successful (see example below). Use induction to show that for n greaterthanorequalto 1, no matter how the dots are arranged on the circle, there is always a place that you can start on the circle so that your trip will be successful. including words to explain what you are doing.

Solution

Given there are 2n dots around the circle, where n dots are red and n are blue

for the inductive step 2k+2 dots are there

in which let us considet the starting point as one dot is blue followed by one red dot .

on removing these two dots By the method of induction hypothesis there is a usable starting point for the reduced

circle.

Then we can check whether that starting point still works when you restoe those two adjacent dots which we

removed earlier.

in our case now there are only 2k dots on removing these two dots. Just check the condition of having count of red

dots more than blue dots.

After that replace those two removed dots where it is 2k+2 dots then again starting from the starting condition

of having count of red dots more than blue dots.

So that we can say that the trip was successful.