The question consists of a combinatorial problem where we are asked to distribute 8 identical balls into 4 distinct small triangles, each triangle must contain at least one ball. Here, any given triangle can contain 1, 2, 3, or more up to 8 balls but not 0.

Option 1:"32" is indeed the correct answer. This utilizes the "Stars and Bars" combinatorial principle, where instead of 8 balls, we consider distributing 8-4 = 4 "extra" balls (as one ball is already in each triangle). The number of ways is thus C(4+4-1, 4) = 35. But, we have to deduct cases in which any triangle gets more than 5 (8-3=5) balls, which is C(4,1)*C(3+4-1,3)=3. Thus, final answer = 35-3=32.

Option 2: "35" signifies the total number of ways 8-4 balls can be distributed in 4 boxes, regardless of any further conditions.

Option 3: "44" has no specific reasoning or computations related to this problem.

Option 4: "56" also