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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