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A set of \(2 N+1\) rods consists of one of each integer length \(1,2, \ldots, 2 N, 2 N+1\) Three, of lengths \(a, b\) and \(c\), are selected, of which \(a\) is the longest. By considering the possible values of \(b\) and \(c\), determine the number of ways in which a nondegenerate triangle (i.e. one of non-zero area) can be formed (i) if \(a\) is even, and (ii) if \(a\) is odd. Combine these results appropriately to determine the total number of non-degenerate triangles that can be formed with the \(2 N+1\) rods, and hence show that the probability that such a triangle can be formed from a random selection (without replacement) of three rods is $$ \frac{(N-1)(4 N+1)}{2\left(4 N^{2}-1\right)} $$

Step by step solution

01

- Understanding the Triangle Inequality

To form a non-degenerate triangle with sides of lengths a, b, and c, where a is the longest side, the triangle inequality must be satisfied: \(a < b + c\).

02

- Evaluating the Maximum for Side a

Given the rod lengths are \(1, 2, \ldots, 2N, 2N+1\), let’s assume \(a\) is the longest side. If \(a = 2k\) (even), the possible scenarios for \(b\) and \(c\) are \(b < a\) and \(c < a\).

03

- Counting the Possibilities for Even a

For \(a = 2k\), we have \(b\) and \(c\) such that \(b + c > 2k\). This implies different cases based on the values \(b\) and \(c\) can take. We'll need to count valid combinations.

04

- Evaluating the Maximum for Side a if Odd

If \(a = 2k+1\) (odd), similar constraints of \(a\) still hold: \(a < b + c\) with \(b < a\) and \(c < a\).

05

- Counting the Possibilities for Odd a

Use similar methods to tally combinations for \(b\) and \(c\) such that \(b + c > 2k+1\) while ensuring both \(b, c \leq 2k\).

06

- Summing Up Even and Odd Cases

Sum combinations for both even and odd \(a\) to find the total number of ways to select three lengths \(a, b, c\) to form non-degenerate triangles.

07

- Calculating Total Possible Selections

The total possible ways to select any 3 rods from \(2N+1\) is \(\binom{2N+1}{3}\).

08

- Finding the Probability

Calculate the probability that a randomly selected set of 3 rods form a non-degenerate triangle by dividing the number of valid triangles by the total possible selections.

09

- Final Expression for Probability

Using above steps, derive and simplify the probability formula to show that it equals \(\frac{(N-1)(4N+1)}{2(4N^2-1)}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Triangle Inequality

To form a non-degenerate triangle (one with a non-zero area), we need to satisfy the triangle inequality condition: for any chosen sides of lengths \(a\), \(b\), and \(c\) with \(a \geq b \geq c\), the following must hold true:
\[ a < b + c \]
This means the length of the longest side must be less than the sum of the other two sides. This rule is important because, without it, the 'triangle' would just be a straight line or not possible at all. When working on problems involving rods or segments, always remember this fundamental rule to determine if a valid triangle can form.

Combinatorics

Combinatorics is a branch of mathematics dealing with counting, combination, and permutation. Here, we use combinatorics to determine the number of ways to select sets of rods that can form a triangle. Given a set of \(2N + 1\) rods with lengths ranging from 1 to \(2N + 1\), we need to count the valid combinations of three rods that satisfy the triangle inequality. To do so:

• Identify all possible combinations of three rods using \(\binom{2N+1}{3}\).
• Filter combinations where the chosen rods meet the triangle inequality.

By considering cases where the longest rod \(a\) is either even or odd, we can simplify the counting process. This allows us to methodically evaluate different scenarios without missing any valid combinations.

Probability Calculation

After determining the number of valid ways to select three rods forming a triangle, we compute the probability. Here is the step-by-step process:

• Calculate the total number of ways to select any three rods from \(2N + 1\) rods using the combinations formula: \(\binom{2N+1}{3}\).
• Determine the number of valid combinations that form a non-degenerate triangle.
• Divide the number of valid combinations by the total combinations to get the probability.

Thus, if \(V\) is the number of valid triangles, and \(T\) is the total combinations, the probability \(P\) is given by:
\[ P = \frac{V}{T} \] For our specific case, the probability formula is simplified to:
\[ \frac{(N-1)(4N+1)}{2(4N^2-1)} \] This step-by-step approach ensures clarity and methodically breaks down the problem.

Integer Length Rods

In this problem, the rods have integer lengths from 1 to \(2N + 1\). The key steps to solve the problem involve understanding how:

• Each rod’s length impacts the formation of a triangle.
• The parity (odd or even nature) of the longest rod affects the combinations.

If the longest rod, \(a\), is even:

• Consider lengths for \(b\) and \(c\) such that \(b + c > a\).
• Evaluate maximum possible pairings without exceeding \(a\).

Similarly, if the longest rod, \(a\), is odd:

• Parameters for \(b\) and \(c\) change slightly but still must satisfy \(b + c > a\).
• Count pairings systematically.

Working through the problem systematically by considering even and odd cases separately helps in managing the complexity and ensuring all valid combinations are counted.