There are two variations I would like to discuss about this problem that I have encountered and haven't been able to solve previously.

1) Shuffling is introduced i.e when order/arrangement matters. Like suppose we have a1 + a2 + a3 = 4 then one of the solutions is (a1,a2,a3) = (1,1,2) , but this should not be counted as 1 arrangement but instead as (1+1+2)! / 2! arrangements because the order of the arrangement matters i.e (a1,a2,a3,a3) is different from (a1,a3,a2,a3). Is there an efficient way to count this ?

2) No shuffling but similar sets should be counted only once. i.e (a1,a2,a3) = (1,1,2) is same as (a1,a2,a3) = (2,1,1)

Also, what is a good blog/site/resource for intermediate-hard counting/combinatorics problems (with editorials/theory).