The number of zeros at the end of 60! is :

Answer: Option B

Clearly, highest power of 2 is much higher as compared to that of 5 in 60!,
So, Required number of zeros
= Highest power of 5
= $$\left[\frac{60}{5}\right]+\left[\frac{60}{5^2}\right]$$
= 12 + 2
= 14