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 $latex 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
In 2026, the following Cheenta students have been successful for Indian Statistical Institute's M.Stat Entrance. They ranked within the first 50 in the entire country in these entrances. I.S.I. M.Stat Entrance
In 2026, the following Cheenta students have been successful for Indian Statistical Institute's B.Stat Entrance and Chennai Mathematical Institute's B.Sc. Math Entrance. They ranked within the first 200 in the entire country in these entrances. Most of these students attended the problem solving workshops regularly, which happen 5 days every week. CMI B.Sc. Math Entrance […]
In 2025, 8 students from Cheenta Academy cracked the prestigious Regional Math Olympiad. In this post, we will share some of their success stories and learning strategies. The Regional Mathematics Olympiad (RMO) and the Indian National Mathematics Olympiad (INMO) are two most important mathematics contests in India.These two contests are for the students who are […]
Cheenta Academy proudly celebrates the success of 27 current and former students who qualified for the Indian Olympiad Qualifier in Mathematics (IOQM) 2025, advancing to the next stage — RMO. This accomplishment highlights their perseverance and Cheenta’s ongoing mission to nurture mathematical excellence and research-oriented learning.