Extremal Principle is used in a variety of problems in Math Olympiad. The following problem from AMC 10 is a very nice example of this idea.

## AMC 10 Problem 4 (2019)- Based on Extremal Principle

A box contains 28 red balls, 20 green balls, 19 yellow balls, 13 blue balls, 11 white balls, and 9 black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least 15 balls of a single color will be drawn?

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Suppose there 5 blue balls, 7 white balls, and 9 green balls. At least how many balls should you pick up (without looking a and without replacement) to be sure that you have picked up at least 4 balls of the same color?

You can also try these problems related to spiral similarity.

You may also click the link to learn its application:- https://www.youtube.com/watch?v=8o8AAWt960o

## 8 replies on “Extremal Principle for Counting – AMC 10”

10 is the answer

That is right

3+3+3+1=10

Correct!

Is the answer 76? My internet is slow and I’m unable to watch the video. According to my analysis it’s 14+14+14+13+11+9=75, which is the most unlucky case possible, hence to ensure 15 balls of a same colour, we need to take out one extra. That gives 76.

Correct

The answer to the successive given question is 3+3+3+(1)=10, as 3+3+3 is the most unluckiest condition.

Yes!