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# Arrangement of Similar Items (TOMATO Subjective 49)

Problem: $$x$$ red balls, $$y$$ black balls,$$z$$ white balls are to be arranged in a row. Suppose that any two balls of the same color are indistinguishable. Given $$x+y+z=30$$ prove that number of possible arrangements is maximum when $$x=y=z=10$$.

Solution: Given a total of $$n$$ objects out of which $$r$$ are of the same type, total number of arrangements possible is given by $$\frac{n!}{r!}$$. Therefore in this case the total number of arrangements is $$\frac{30!}{x!y!z!}$$.

Now we have to maximise $$\frac{30!}{x!y!z!}$$.

Thus we need to minimise $${x!y!z!}$$. As the question claims that all have to be equal to $$10$$, let us consider that they are not all equal to $$10$$. So the cases that arise are:

November 6, 2016