There 52 balls in a box: 16 red, 16, blue, 16 yellow, and 4 that are either black or white. What is the least number of balls you must take out of the box (Without looking) to be sure that amount them there are 8 balls of the same color?
The Correct Answer and Explanation is :
To determine the least number of balls you must take out of the box to ensure that there are at least 8 balls of the same color, we can apply the pigeonhole principle. The principle states that if ( n ) items are put into ( m ) containers, with ( n > m ), then at least one container must contain more than one item.
Step-by-Step Analysis
- Understanding the Colors:
- There are three colors with 16 balls each: red, blue, and yellow.
- There are 4 balls that are either black or white, but we will not use these colors since we are focused on obtaining 8 balls of the same color.
- Applying the Pigeonhole Principle:
- We want to ensure we have 8 balls of at least one color.
- We can consider the worst-case scenario, where we take out the maximum number of balls without getting 8 of the same color.
- Calculating Maximum Balls Without Reaching 8:
- If we draw 7 balls of each of the three colors (red, blue, and yellow), that totals:
[
7 \text{ (red)} + 7 \text{ (blue)} + 7 \text{ (yellow)} = 21 \text{ balls}
] - At this point, we have not yet reached 8 balls of any color.
- Drawing More Balls:
- Now, if we draw one more ball (the 22nd ball), regardless of the color, it will have to be either red, blue, or yellow. This will increase the count of that color to 8.
Conclusion
Thus, to ensure that there are at least 8 balls of the same color, you need to take out 22 balls.
This logic demonstrates how the pigeonhole principle can effectively solve problems involving distributions and guarantees of quantities in combinatorial settings. By strategically considering the worst-case scenario, we can ensure we meet our criteria with the least effort and maximum efficiency.