Key Concepts and Formulas

• Stars and Bars: The number of ways to distribute $n$ identical items into $k$ distinct bins, where each bin can have any number of items (including zero), is given by ${}^{n+k-1}{C_{k-1}}$.
• Stars and Bars with Minimum Requirement: The number of ways to distribute $n$ identical items into $k$ distinct bins such that each bin receives at least one item is given by ${}^{n-1}{C_{k-1}}$.
• Combinations: The number of ways to choose $r$ items from a set of $n$ items is given by ${}^n{C_r} = \frac{n!}{r!(n-r)!}$.

Step-by-Step Solution

• Step 1: Identify the Problem Type and Given Values. We are given the problem of distributing $n = 8$ identical balls into $k = 3$ distinct boxes, with the condition that none of the boxes is empty. This is a classic "Stars and Bars" problem with a minimum requirement for each box.

• Step 2: Apply the Stars and Bars Formula with Minimum Requirement. Since each box must contain at least one ball, we use the formula ${}^{n-1}{C_{k-1}}$, where $n = 8$ and $k = 3$. Number of ways = ${}^{8-1}{C_{3-1}} = {}^{7}{C_2}$. Why this formula? The formula ${}^{n-1}{C_{k-1}}$ directly accounts for the constraint that each box must have at least one ball. It's equivalent to first placing one ball in each of the $k$ boxes, and then distributing the remaining $n-k$ balls without restriction.

• Step 3: Calculate the Combination. We calculate ${}^{7}{C_2}$. ${}^{7}{C_2} = \frac{7!}{2!(7-2)!} = \frac{7!}{2!5!} = \frac{7 \times 6 \times 5!}{2 \times 1 \times 5!} = \frac{7 \times 6}{2} = 7 \times 3 = 21$. Why this calculation? We are using the definition of combinations to find the number of ways to choose 2 items from a set of 7 items.

Common Mistakes & Tips

• Confusing Formulas: Be careful to use the correct Stars and Bars formula based on whether empty bins are allowed or not. If empty bins are allowed, use ${}^{n+k-1}{C_{k-1}}$. If empty bins are not allowed, use ${}^{n-1}{C_{k-1}}$.
• Identical vs Distinct: Always identify whether the items and bins are identical or distinct. This problem involves identical items (balls) and distinct bins (boxes), making Stars and Bars the appropriate technique.
• Alternative Approach: This problem can be viewed as finding the number of positive integer solutions to the equation $x_1 + x_2 + x_3 = 8$. This is equivalent to the Stars and Bars problem with the minimum requirement.

Summary

The problem involves distributing 8 identical balls into 3 distinct boxes such that none of the boxes is empty. Using the Stars and Bars formula with a minimum requirement, the number of ways is calculated as ${}^{n-1}{C_{k-1}} = {}^{8-1}{C_{3-1}} = {}^{7}{C_2} = 21$.

Final Answer The final answer is \boxed{21}, which corresponds to option (B).