There's a formula for finding the power of a prime p in n!.
If n!=pa.qbrc...
Then

(where [.] represents greatest integer function)
similarly for q,b and r,c and all other prime factors of n!..
To find no. of zeroes in 2002C1001, you must know power of 5 in 2002! and 1001! since
2002C1001 
So simply find the power of 5 and that will give you the no. of zeroes...