Key Concepts and Formulas

• Stars and Bars Method: The number of non-negative integer solutions to $x_1 + x_2 + \dots + x_k = n$ is given by $\binom{n+k-1}{k-1} = \binom{n+k-1}{n}$.
• Lower Bound Constraints: If $x_i \ge a$, substitute $y_i = x_i - a$ to obtain a non-negative variable $y_i \ge 0$.
• Principle of Inclusion-Exclusion (PIE): For two conditions $A$ and $B$, the number of solutions satisfying neither is $N(\text{Total}) - [N(A) + N(B)] + N(A \cap B)$.

Step-by-Step Solution

Step 1: Problem Setup and Initial Equation

We want to distribute 30 identical candies among four children $C_1, C_2, C_3,$ and $C_4$. Let $x_1, x_2, x_3, x_4$ represent the number of candies each child receives, respectively. We are given the equation: $x_1 + x_2 + x_3 + x_4 = 30$ with the constraints:

• $x_1 \ge 0$
• $4 \le x_2 \le 7$
• $2 \le x_3 \le 6$
• $x_4 \ge 0$

Step 2: Handling Lower Bound Constraints

To apply Stars and Bars, we need all variables to be non-negative. We handle the lower bounds for $x_2$ and $x_3$ by substitution:

• Let $y_2 = x_2 - 4$. Then $x_2 = y_2 + 4$ and $y_2 \ge 0$.
• Let $y_3 = x_3 - 2$. Then $x_3 = y_3 + 2$ and $y_3 \ge 0$.

Substituting these into the equation gives: $x_1 + (y_2 + 4) + (y_3 + 2) + x_4 = 30$ $x_1 + y_2 + y_3 + x_4 = 30 - 4 - 2 = 24$ Now we have $x_1 + y_2 + y_3 + x_4 = 24$ with $x_1, y_2, y_3, x_4 \ge 0$.

We also need to update the upper bound constraints:

• $x_2 \le 7 \Rightarrow y_2 + 4 \le 7 \Rightarrow y_2 \le 3$
• $x_3 \le 6 \Rightarrow y_3 + 2 \le 6 \Rightarrow y_3 \le 4$

So now we have $x_1 + y_2 + y_3 + x_4 = 24$ with constraints $y_2 \le 3$ and $y_3 \le 4$.

Step 3: Calculate Total Solutions Without Upper Bounds

Ignoring the upper bounds $y_2 \le 3$ and $y_3 \le 4$, the number of non-negative integer solutions to $x_1 + y_2 + y_3 + x_4 = 24$ is: $\binom{24 + 4 - 1}{4 - 1} = \binom{27}{3} = \frac{27 \times 26 \times 25}{3 \times 2 \times 1} = 2925$

Step 4: Applying the Principle of Inclusion-Exclusion

Let $A$ be the set of solutions where $y_2 \ge 4$ (violating $y_2 \le 3$), and $B$ be the set of solutions where $y_3 \ge 5$ (violating $y_3 \le 4$). We want to find the number of solutions that satisfy neither $A$ nor $B$. By PIE: $N(\text{Desired}) = N(\text{Total}) - [N(A) + N(B)] + N(A \cap B)$

Step 5: Calculate N(A)

If $y_2 \ge 4$, let $z_2 = y_2 - 4$. Then $y_2 = z_2 + 4$ and $z_2 \ge 0$. Substituting into $x_1 + y_2 + y_3 + x_4 = 24$: $x_1 + (z_2 + 4) + y_3 + x_4 = 24$ $x_1 + z_2 + y_3 + x_4 = 20$ The number of solutions is $\binom{20 + 4 - 1}{4 - 1} = \binom{23}{3} = \frac{23 \times 22 \times 21}{3 \times 2 \times 1} = 1771$. So $N(A) = 1771$.

Step 6: Calculate N(B)

If $y_3 \ge 5$, let $z_3 = y_3 - 5$. Then $y_3 = z_3 + 5$ and $z_3 \ge 0$. Substituting into $x_1 + y_2 + y_3 + x_4 = 24$: $x_1 + y_2 + (z_3 + 5) + x_4 = 24$ $x_1 + y_2 + z_3 + x_4 = 19$ The number of solutions is $\binom{19 + 4 - 1}{4 - 1} = \binom{22}{3} = \frac{22 \times 21 \times 20}{3 \times 2 \times 1} = 1540$. So $N(B) = 1540$.

Step 7: Calculate N(A ∩ B)

If $y_2 \ge 4$ and $y_3 \ge 5$, let $z_2 = y_2 - 4$ and $z_3 = y_3 - 5$. Then $y_2 = z_2 + 4$ and $y_3 = z_3 + 5$. Substituting into $x_1 + y_2 + y_3 + x_4 = 24$: $x_1 + (z_2 + 4) + (z_3 + 5) + x_4 = 24$ $x_1 + z_2 + z_3 + x_4 = 15$ The number of solutions is $\binom{15 + 4 - 1}{4 - 1} = \binom{18}{3} = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = 816$. So $N(A \cap B) = 816$.

Step 8: Final Calculation

Using PIE: $N(\text{Desired}) = 2925 - (1771 + 1540) + 816 = 2925 - 3311 + 816 = 430$

Common Mistakes & Tips

• Always make the variables non-negative before applying Stars and Bars.
• Remember to adjust upper bounds after making substitutions for lower bounds.
• Be careful with the signs in the Principle of Inclusion-Exclusion.

Summary

By applying the Stars and Bars method and the Principle of Inclusion-Exclusion, we found the number of ways to distribute the candies according to the given constraints. The final answer is 430.

Final Answer

The final answer is \boxed{430}, which corresponds to option (D).