To find trailing zeros in $n!$, sum $\lfloor n/5^k \rfloor$. For $123!$: $\lfloor 123/5 \rfloor = 24$; $\lfloor 123/25 \rfloor = 4$. Total zeros = $24 + 4 = 28$. Options like 24 or 25 are incorrect because they miss the higher powers of 5 (like 25, 50, 75, 100).