Question 397124: Hello,
I have this math brain teaser that is killing me. Can you help please?
In a dark room, a boxed is filled with different colored balls there are 14 blue, 13 red, 22 black and 18 green. What is the smallest number of balls you need to select to get at least 4 of the same color.
Kindest regards,
Candy
Found 2 solutions by solver91311, richard1234:
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
If you choose 12 of the balls, it is possible to select 3 of each of the 4 colors. Then if you take one more, you will have 4 of whichever color you selected on the 13th ball. So, 13.
John
My calculator said it, I believe it, that settles it
Answer by richard1234(7193)  (Show Source):
You can put this solution on YOUR website! Note that the Pigeonhole Principle says that if there if we have kn + 1 pigeons distributed into n holes, then there is at least one hole with k+1 pigeons. In this case, n = 4 (there are four colors, or "holes"), and we want to guarantee that one of these "holes" has four "pigeons." Therefore, letting k = 3, we see that if we draw 3(4) + 1 or 13 balls, we are guaranteed to have four of the same color.
If there are 12 balls, then it is possible to draw three of each color.
|
|
|
