The number of ways, 16 identical cubes, of which 11 are blue and rest are red, can be placed in a ro

Q: Wait, why do we have 6 gaps for 5 red cubes? Shouldn't it be 5 gaps?

That is a classic 'fencepost' error! Think about it: if you have 5 red cubes in a row, say $R R R R R$, you have a space before the first one, a space after the last one, and 4 spaces between the 5 cubes. That is $1 + 4 + 1 = 6$ total spaces. Always draw it out if you are unsure!

Q: Why can $x_1$ and $x_6$ be 0, but the middle ones must be at least 2?

The problem specifically states the constraint applies 'between any two red cubes.' It does not say there must be blue cubes at the ends. Therefore, the ends are free to be empty ($0$ blue cubes), while the middle gaps are strictly bound by the 'at least 2' rule.

Q: How does the substitution $x_i = t_i + 2$ actually help?

The 'Stars and Bars' formula, $\binom{n+r-1}{r-1}$, only works for non-negative integers ($x_i \ge 0$). By setting $x_i = t_i + 2$, we ensure that even if $t_i$ is $0$ (the minimum possible value for $t_i$), $x_i$ becomes $2$, which is the minimum required by your constraint. It effectively 'locks in' the required cubes so you can distribute the remaining ones freely.

Question

The number of ways, 16 identical cubes, of which 11 are blue and rest are red, can be placed in a row so that between any two red cubes there should be at least 2 blue cubes, is ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a row of sixteen identical cubes. Eleven are a deep, ocean blue, and five are a fiery red. Your task is to arrange them in a single line, subject to the constraint that between any two red cubes, there must be at least two blue cubes.

### The Gap Method: Visualizing the Structure

To solve this, we visualize the structure created by the red cubes. If we place our 5 red cubes in a row, we create gaps where the blue cubes can be placed: $\_ R \_ R \_ R \_ R \_ R \_$.

There is a gap before the first red cube, a gap after the last red cube, and gaps between each consecutive pair of red cubes. Counting these, we find exactly 6 gaps.

Let $x_1$ be the number of blue cubes before the first red, $x_2, x_3, x_4, x_5$ be the blue cubes in the middle gaps, and $x_6$ be the blue cubes after the last red. Since we have 11 blue cubes in total, the sum of these variables must satisfy:

$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 11$$

### The Constraint: Pre-allocation

The problem demands at least 2 blue cubes between any two red cubes. This implies the constraints $x_2, x_3, x_4, x_5 \ge 2$, while $x_1$ and $x_6$ have no such restriction ($x_1, x_6 \ge 0$).

To apply the 'Stars and Bars' theorem, we perform a substitution for the middle gaps: $x_i = t_i + 2$ for $i = 2, 3, 4, 5$, where $t_i \ge 0$. By pre-allocating 2 blue cubes to each of the 4 middle gaps, we use up $4 	imes 2 = 8$ blue cubes.

The equation transforms as follows:

$$x_1 + (t_2 + 2) + (t_3 + 2) + (t_4 + 2) + (t_5 + 2) + x_6 = 11$$

Simplifying this, we obtain:

$$x_1 + t_2 + t_3 + t_4 + t_5 + x_6 = 3$$

### The Final Calculation: Stars and Bars

We are now left with 3 blue cubes to distribute freely among 6 gaps. The number of non-negative integer solutions to $\sum_{i=1}^{r} a_i = n$ is given by the formula $\binom{n+r-1}{r-1}$.

Here, $n = 3$ and $r = 6$. Plugging these values into the formula, we get:

$$\binom{3+6-1}{6-1} = \binom{8}{5}$$

Using the symmetry property of combinations, $\binom{8}{5} = \binom{8}{3}$. Calculating this value:

$$\binom{8}{3} = \frac{8 	imes 7 	imes 6}{3 	imes 2 	imes 1} = 56$$

Through the elegance of logical constraints and combinatorial substitution, we have determined that there are exactly **56** ways to arrange these cubes.

Animated Solution for Mathematics - Permutations and Combinations: The number of ways, 16 identical cubes, of which 11 are blue and rest are red, can be placed in a row so that between any two red cubes there should be at least 2 blue cubes, is ______.

Enter Numerical Value:

Visualized Solution

Analyze the Given Cubes

Define the Variables for Gaps

Formulate the Main Equation

Apply the Constraints

Variable Substitution

Substitute into the Equation

Simplify the Equation

Apply the Multinomial Formula

Calculate the Combinations

Final Answer

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Gap Method

The Constraint

The Final Calculation

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