There is a straightforward method to solve this type of question. If we split $200!$ into its prime factors, then we will end up with many $2$s and $5$s. It is these prime factors that are going to give us factors of $10$ which in turn give us trailing zeros. There are going to be many more $2$s than $5$s, since we will get a prime factor of $2$ for every even factor. Therefore, we just have to count the number of $5$s. $$ $$ We have to be careful here, as $25$ and its multiples will contribute two $5$s each as their prime factors contain $5^2$, and $125$ will contribute three $5$s as its prime factors are $5^3$. Fortunately, we can simply double count here to take these into account. So, the procedure is as follows: count all multiples of $5$ less than or equal to $200$, and then all multiples of $25$, and then all multiples of $125$. Add these counts together. $$ $$ We get $40$ multiples of $5$ $(5, 10, 15, 20...)$, $8$ multiples of $25$ $(25, 50, 75, 100...)$ and $1$ multiple of $125$. Adding these together gives us our result - there are $49$ trailing zeroes in $200!$.