I'm trying to prove the following statement:
Suppose that you have 25 balls to place into five different bins. 11 of the balls are red, while the other 14 are blue. Prove that no matter how the balls are placed into the bins, there must be at least one bin containing at least 3 red balls.
My proof is the following: By contradiction. Assume that it's possible to place 25 balls into 5 bins in the way that there is no a bin with 3 or more red balls. Consider any such way of placing the balls, and let the number of red balls in the five bins be a, b, c, d, e. Then, the maximum number of red balls in five bins will be 10, because we assumed that there is no a bin with 3 or more red balls. Since we have 11 red balls that have to be placed, we have reached a contradiction, so our initial assumption must have been wrong and there must be at least one bin with at least 3 red balls.
Is my proof correct or not? I'm asking because I'm worrying why there are so many numbers in the statement (25 balls, 14 blue balls) if the fact that there is only 11 red balls is enough to show that it's impossible to place 11 red balls into five bins and to have less than 3 balls in each bin.