I read a proof about the simple random walk in 3 dimensions and couldn't understand the following statement: $$\frac{n!}{k!j!(n-k-j)!}$$ has the maximum when $\ k, j $ and $\ n-k-j$ are as close to $\frac{n}{3}$ as possible.
Does anyone show me a proof for this?
That is equivalent to show how to minimize $a!b!c!$ with $a+b+c=n$. Suppose $a-b>1$, then replace $a,b$ by $a-1,b+1$ we will get a smaller product as $(a-1)b(b+1)<(a-1)ab$. By the symmetry of $a,b,c$, the result follows, since $a,b,c$ are as close to $n/3$ as possible means exactly each pair in $a,b,c$ has no differnce greater than $1$.