Solution:
The number of zeroes at the end of 60! depends on the number of times 60! is a multiple of 10.
Now 10 = 2*5.
So what we need to do is to find out how many pairs of 2"s and 5's are contained in 60!.
Since every second integer in 60! = 1*2*3*4...*59*60 is a multiple of 2, we only need to calculate the number of 5's in 60!.
Each of the 5's will combine with a 2 to give 10.

Multiples of 5 in 60! are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60. They are 12 in number but 25 = 5^2, and so it has two 5's and 50 = (5^2)*2 , which again has two 5's.
So in total there are 12+2 = 14 fives.
Or 60! will have 14 zeroes in the end.