Question

{"given":"We have 8 identical balls to be distributed among 3 distinct boxes. The constraint is that none of the boxes should remain empty. This is a classic combinatorics problem involving distribution of identical objects.","key_observation":"This is a 'stars and bars' problem with restrictions. Since no box can be empty, we first place one ball in each of the 3 boxes (using 3 balls). Then we need to distribute the remaining 5 balls among the 3 boxes without any restrictions. The formula becomes $\\binom{n-1}{r-1}$ where $n$ is the remaining balls and $r$ is the number of boxes.","option_analysis":[{"label":"(A)","text":"5","verdict":"incorrect","explanation":"This value is too small. It doesn't account for the proper combinatorial calculation required for distributing identical objects among distinct boxes."},{"label":"(B)","text":"$\\binom{8}{3}$","verdict":"incorrect","explanation":"This equals 56, which represents choosing 3 positions from 8, but this doesn't solve our distribution problem. The formula doesn't account for the constraint of non-empty boxes."},{"label":"(C)","text":"$3^8$","verdict":"incorrect","explanation":"This equals 6561, which would be the number of ways to distribute 8 distinct balls among 3 boxes (each ball has 3 choices). However, our balls are identical, not distinct."},{"label":"(D)","text":"21","verdict":"correct","explanation":"Using stars and bars with restriction: place 1 ball in each box first, then distribute remaining 5 balls among 3 boxes. This gives $\\binom{5+3-1}{3-1} = \\binom{7}{2} = \\frac{7 \\times 6}{2} = 21$."}],"answer":"(D)","formula_steps":[]}