Key Concepts and Formulas

• Stars and Bars: The number of non-negative integer solutions to $x_1 + x_2 + \dots + x_n = k$ is given by $\binom{k + n - 1}{n - 1} = \binom{k + n - 1}{k}$.
• Transforming Variables: If a variable $x_i$ has a minimum value constraint $x_i \ge m$, we can transform it into a non-negative variable $y_i$ by setting $x_i = y_i + m$, where $y_i \ge 0$.
• Combinations: The number of ways to choose $k$ items from a set of $n$ items is given by $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Step-by-Step Solution

Step 1: Problem Setup and Variable Definition

We have 11 blue cubes (B) and 5 red cubes (R) to arrange in a row such that there are at least 2 blue cubes between any two red cubes. We want to find the number of such arrangements. We can represent the arrangement as:

$\underbrace{B \ B \ ...}_{x_1} \ R \ \underbrace{B \ B \ ...}_{x_2} \ R \ \underbrace{B \ B \ ...}_{x_3} \ R \ \underbrace{B \ B \ ...}_{x_4} \ R \ \underbrace{B \ B \ ...}_{x_5} \ R \ \underbrace{B \ B \ ...}_{x_6}$

Here, $x_i$ represents the number of blue cubes in each region:

• $x_1$: Before the first red cube ($x_1 \ge 0$).
• $x_2$: Between the first and second red cubes ($x_2 \ge 2$).
• $x_3$: Between the second and third red cubes ($x_3 \ge 2$).
• $x_4$: Between the third and fourth red cubes ($x_4 \ge 2$).
• $x_5$: Between the fourth and fifth red cubes ($x_5 \ge 2$).
• $x_6$: After the fifth red cube ($x_6 \ge 0$).

The sum of these variables must equal the total number of blue cubes, which is 11.

Step 2: Formulating the Initial Equation

Since we have 11 blue cubes, the sum of blue cubes in all regions must equal 11: $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 11$

Step 3: Incorporating the Constraints

We have the constraints $x_1 \ge 0$, $x_2 \ge 2$, $x_3 \ge 2$, $x_4 \ge 2$, $x_5 \ge 2$, and $x_6 \ge 0$. To apply Stars and Bars, we need all variables to be non-negative. We'll transform the variables $x_2, x_3, x_4,$ and $x_5$.

Step 4: Transforming the Variables

Let's define new variables:

• $y_1 = x_1$
• $y_2 = x_2 - 2 \implies x_2 = y_2 + 2$
• $y_3 = x_3 - 2 \implies x_3 = y_3 + 2$
• $y_4 = x_4 - 2 \implies x_4 = y_4 + 2$
• $y_5 = x_5 - 2 \implies x_5 = y_5 + 2$
• $y_6 = x_6$

Substituting these into our equation: $y_1 + (y_2 + 2) + (y_3 + 2) + (y_4 + 2) + (y_5 + 2) + y_6 = 11$ $y_1 + y_2 + y_3 + y_4 + y_5 + y_6 + 8 = 11$ $y_1 + y_2 + y_3 + y_4 + y_5 + y_6 = 3$ Now all $y_i \ge 0$, so we can apply Stars and Bars.

Step 5: Applying Stars and Bars

We have the equation $y_1 + y_2 + y_3 + y_4 + y_5 + y_6 = 3$, where all $y_i$ are non-negative integers. We have $k = 3$ and $n = 6$. Using the Stars and Bars formula, the number of solutions is: $\binom{k + n - 1}{k} = \binom{3 + 6 - 1}{3} = \binom{8}{3}$ $\binom{8}{3} = \frac{8!}{3!5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 8 \times 7 = 56$

Common Mistakes & Tips

• Forgetting to Transform: The most common error is not transforming the variables to be non-negative before applying Stars and Bars.
• Misinterpreting the Constraints: Carefully analyze what "between" means in the problem and which variables are affected by the constraints.
• Calculation Errors: Double-check your combination calculations to avoid simple arithmetic mistakes.

Summary

We used Stars and Bars to find the number of ways to arrange 11 blue cubes and 5 red cubes with the given constraint. We first defined variables to represent the number of blue cubes in different regions. We then transformed the variables to account for the minimum number of blue cubes required between red cubes. Finally, we applied the Stars and Bars formula to calculate the number of possible arrangements, which is 56.

Final Answer

The final answer is \boxed{56}.