Key Concepts and Formulas

• Stars and Bars: The number of non-negative integer solutions to $x_1 + x_2 + \dots + x_k = n$ is $\binom{n+k-1}{k-1} = \binom{n+k-1}{n}$.
• Pre-distribution: To handle the constraint $x_i \ge m$, substitute $x_i = y_i + m$, where $y_i \ge 0$.
• Binomial Coefficient: $\binom{n}{k} = \frac{n!}{k!(n-k)!}$

Step-by-Step Solution

Step 1: Define variables and state the problem's equation. We are distributing 21 identical apples among 3 children. Let $x_1$, $x_2$, and $x_3$ represent the number of apples each child receives, respectively. The total number of apples is 21, so we have the equation: $x_1 + x_2 + x_3 = 21$

Step 2: State the constraints. Each child must receive at least 2 apples. This gives us the constraints: $x_1 \ge 2, \quad x_2 \ge 2, \quad x_3 \ge 2$ The standard Stars and Bars formula requires non-negative variables. Since the variables must be greater than or equal to 2, we need to transform the variables.

Step 3: Transform the variables using pre-distribution. To satisfy the minimum requirement of 2 apples for each child, we pre-distribute 2 apples to each child. Let's define new variables $y_1, y_2, y_3$ such that: $x_1 = y_1 + 2, \quad x_2 = y_2 + 2, \quad x_3 = y_3 + 2$ Substituting these into the original equation ensures that $y_1, y_2, y_3 \ge 0$, which is required for the Stars and Bars formula.

Step 4: Substitute the transformed variables into the equation. Substituting the expressions for $x_1, x_2, x_3$ into the equation $x_1 + x_2 + x_3 = 21$, we get: $(y_1 + 2) + (y_2 + 2) + (y_3 + 2) = 21$ $y_1 + y_2 + y_3 + 6 = 21$

Step 5: Simplify the equation. Subtract 6 from both sides of the equation: $y_1 + y_2 + y_3 = 15$ Now we have an equation with non-negative integer solutions, suitable for applying Stars and Bars.

Step 6: Apply the Stars and Bars formula. We want to find the number of non-negative integer solutions to $y_1 + y_2 + y_3 = 15$. Here, $n = 15$ and $k = 3$. Using the Stars and Bars formula, the number of solutions is: $\binom{n + k - 1}{k - 1} = \binom{15 + 3 - 1}{3 - 1} = \binom{17}{2}$

Step 7: Calculate the binomial coefficient. $\binom{17}{2} = \frac{17!}{2!(17-2)!} = \frac{17!}{2!15!} = \frac{17 \times 16}{2 \times 1} = 17 \times 8 = 136$

Common Mistakes & Tips

• Failing to account for the minimum requirement. Applying Stars and Bars directly without pre-distributing will lead to an incorrect answer.
• Incorrectly calculating the binomial coefficient. Make sure to cancel out the factorials properly.
• Forgetting that Stars and Bars only applies when distributing identical items into distinct containers.

Summary

To solve this problem, we first recognized that it involved distributing identical items (apples) into distinct containers (children) with a minimum requirement for each container. We used the pre-distribution technique to transform the variables and satisfy the minimum requirement, allowing us to apply the Stars and Bars formula. Finally, we calculated the binomial coefficient to arrive at the number of ways to distribute the apples.

Final Answer The final answer is \boxed{136}, which corresponds to option (B).