Step 1: Prime-factorise and count distributions.
\(24=2^3\cdot 3^1\). Distribute powers of 2 among \(x,y,z\): solutions to \(a+b+c=3\) with \(a,b,c\geq0\) is \(\binom{5}{2}=10\). Distribute power of 3: solutions to \(a+b+c=1\) is \(\binom{3}{2}=3\).
Step 2: Multiply the counts.
Total ordered triples \(=10\times3=30\). \[ \boxed{30} \]