@sumit_agarwal I would urge you to consider a few examples yourself and try and work out first along the lines of above logic. Do it by standard approach first. It would help you develop a more generic intellect towards non-standardised questions like these. Feel free to comment again if it doesn't work out. Will try and provide a solution for your query then.

Sum of all elements of set S = 465.
For any subset having sum of elements x, there will be one complimentary set which is having sum of elements equal to (465 – x).
Example let there be a set {1, 2, 3, 4}.
Then subset which is having sum =3 will be {1, 2} {3}: (2)
and subset having sum (1+2+3+4) -3 =7 will also be 2: {3,4} {1,2,4}
So, number of subsets having sum of elements x = Number of subsets having sum of elements (465 - x)
=> No of subsets having sum 1 = No of subsets having sum 464
Also, No of subsets having sum 2 = No of subsets having sum 463
….
No of subsets having sum 232 = No of subsets having sum 233
Adding them we will get,
No of subsets having sum less than or equal to 232
= No of subsets having sum more than 233
= 2^30/2
= 2^29

42 = 2 * 3 * 7
No of zeroes in 2006! will be the highest power of 7 in 2006!
[2006/7] + [2006/49] + [2006/343] = 331
as 42 in base 42 will be 10 and will be the provider of trailing zeroes

In keynote 4, example 2, total non-negative unordered integral solutions is just 1 (4,0) and total non-negative ordered integral solutions are 2 (4,0 and 0,4). Please Correct me if I'm wrong.