INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More 

March 7, 2020

Maximizing Arrangements

x red balls, y black balls and z white balls are to be arranged in a row. Suppose that any two balls of the same color are indistinguishable. Given that x+y+z=30, show that the number of possible arrangements is the largest for x=y=z=10.

Solution: The total number of ways of arranging the 30 balls are frac {(x+y+z)!}{x! y! z!} = frac {30!}{x! y! z!} . Thus we can maximize the number of arrangements by minimizing the denominator. Our claim is that the denominator is least when x=y=z=10.

Suppose otherwise. Let x=10+m , y=10-n, z=10-r minimize the denominator (since all are not equal there will be at least one which is larger than 10; without loss of generality we assume that to be

Other useful links:-

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.