Solution: As far as the question on trailing zero is concerned, there is a direct formal to it.
The number of trailing zero in n! can be calculated through: \(\frac{n}{5} + \frac{n}{5^2} + \frac{n}{5^3} + .... \frac{n}{n^k}\) where \(5^{k+1} > n\)
So calculating by formula, the number of trailing zero in \(50! = \frac{50}{5} + \frac{50}{25} = 10 + 2 = 12\)
However, if you want to understand this logic behind then here is the second method:
Alternate Solution: We know \(50! = 50 \times 49 \times 48 \times 47 \times ....... \times 1\)
Every \(0\) will come from \(2\times 5\). So we basically need to calculate how many \(2\times 5\) do we have in \(50!\).
We can see that there are more \(2's\) than \(5's\) therefore if we calculate the number of \(5's\) we can get the answer.
So the number of multiples of \(5\) between \(1\) to \(50\) \(= \frac{50}{5} = 10\)
So the number of multiples of \(25\) between \(1\) to \(50\) \(= \frac{25}{5} = 2\)
Thus the right answer \(= 10+2 = 12\)
Hence the right answer is
Option E.