Total ways = (Each box has at least 1 copy each) + (One box has 0 copy, Other 3 have at least 1 copy each) + (Two boxes have 0 copy, Other 2 have at least 1 copy each) + (Three boxes have 0 copy, and the last one has all 6 copies)
CASE-1: (Each box has at least 1 copy)
We are left with only 2 balls to distribute because we have distributed one ball to each of the boxes. Can be (1,1,0,0) or (2,0,0,0). [2 WAYS]
NOTE: (1,1,0,0) is same as (1,0,0,1), since boxes are identical
CASE-2: (One box has 0 copy, Other 3 have at least 1 copy each)
We are left with 3 balls to distribute. Can be (0,1,1,1) or (0,1,2,0) or (0,3,0,0). [3 WAYS]
CASE-3: (Two boxes have 0 copy, Other 2 have at least 1 copy each)
We are left with 4 balls to distribute. Can be (0,0,2,2) or (0,0,3,1) or (0,0,4,0). [3 WAYS]
CASE-4: (Three boxes have 0 copy, last one has all the copies)
Only one way (6,0,0,0) [1 WAY]
Therefore total =2+3+3+1=9 ways
If this comes as NAT type question, then remember this can’t be solved as “normal ball-bin problem” because here balls, as well as bins, are identical.