**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:

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