$x$ red balls,$y$ black balls and $z$ white balls are to be arranged in a row.Same coloured balls are indistinguishable and $x+y+z=30$ $x$ red balls,$y$ black balls and $z$ white balls are to be arranged in a row.Suppose that any two balls of the same colour and indistinguishable. Given that $x+y+z=30$,show that the number of possible arrangements is the largest for $x=y=z=10$.  
I know that when $n$ is even,  $n\choose k$ reaches its highest when $k$ is $\frac{n}{2}$.This case too looks similar but I cannot approach it.I can see the result but cannot prove it.Please help!
 A: First place the white balls. That can be done in $\binom{30}{z}$ ways. Then place the black balls. That can be done in $\binom{30-z}{y}$ ways, since there are $30-z$ spots left for them. The red balls can now be placed one way only. All in all we get the total number of ways is
$$
\binom{30}{z} \cdot \binom{30-z}{y}\cdot 1 = \frac{30!}{z!(30-z)!}\cdot \frac{(30-z)!}{y!(30-z-y)!} = \frac{30!}{z!y!(30-(z + y))!}
$$
so it's clear that to get this number as large as possible we need to minimize the denominator (the numerator is fixed).
Now, imagine you're sitting there with the balls in some colour combination other than $10$ of each (say $6, 11, 13$), and write out the denominator $z!y!(30-(z + y))!$ like this (each line representing the contribution from the balls of one colour):
$$
2\cdot3\cdot4\cdot5\cdot6\cdot\\\\2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11\cdot\\\\2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11\cdot12\cdot13
$$
it is clear that if we move one ball (i.e. repaint it) from the lower category into the upper, we turn a $13$ into a $7$, and the total product gets smaller. You can do this to make it smaller until you have all tens, so $10$ of each must give a minimized $z!y!(30-(z + y))!$, hence a maximized number of placements.
