Problem 8 - A Complete Resource Book in Mathematics for JEE Main

Question

Given five line segments of lengths \(2,3,4,5,6\) units. Then the number of triangles that can be formed by joining these lines is (A) \({ }^{5} C_{3}-3\) (B) \({ }^{5} C_{3}-1\) (C) \({ }^{5} C_{3}\) (D) \({ }^{3} C_{3}-2\)

Step-by-Step Solution

Answer

The answer is (A) \(^5C_3 - 3\).

Step 1: Understand the Problem

We need to determine how many triangles can be formed using any three of the five given line segments with lengths 2, 3, 4, 5, and 6 units. We will use the triangle inequality theorem to validate each combination.

Step 2: Calculate Total Combinations

First, find the total number of ways to pick any 3 line segments from 5. This can be calculated using the combination formula \( ^5C_3 \). \( ^5C_3 = \frac{5!}{3! \, 2!} = 10 \).

Step 3: Apply Triangle Inequality Theorem

The triangle inequality theorem states that for any three sides \(a\), \(b\), and \(c\), the sum of any two sides must be greater than the third side. We need to check each of the 10 combinations of 3 side lengths.

Step 4: Evaluate Each Combination

1. (2, 3, 4): \(2 + 3 > 4\), \(2 + 4 > 3\), \(3 + 4 > 2\) - Valid triangle2. (2, 3, 5): \(2 + 3 ot> 5\) - Invalid3. (2, 3, 6): \(2 + 3 ot> 6\) - Invalid4. (2, 4, 5): \(2 + 4 > 5\), \(2 + 5 > 4\), \(4 + 5 > 2\) - Valid triangle5. (2, 4, 6): \(2 + 4 = 6\) - Invalid6. (2, 5, 6): \(2 + 5 > 6\), \(2 + 6 > 5\), \(5 + 6 > 2\) - Valid triangle7. (3, 4, 5): \(3 + 4 > 5\), \(3 + 5 > 4\), \(4 + 5 > 3\) - Valid triangle8. (3, 4, 6): \(3 + 4 > 6\), \(3 + 6 > 4\), \(4 + 6 > 3\) - Valid triangle9. (3, 5, 6): \(3 + 5 > 6\), \(3 + 6 > 5\), \(5 + 6 > 3\) - Valid triangle10. (4, 5, 6): \(4 + 5 > 6\), \(4 + 6 > 5\), \(5 + 6 > 4\) - Valid triangle

Step 5: Count the Valid Triangles

Count the number of valid triangles from the list in the previous step. There are 7 valid combinations: (2, 3, 4), (2, 4, 5), (2, 5, 6), (3, 4, 5), (3, 4, 6), (3, 5, 6), and (4, 5, 6).

Step 6: Match the Answer

From the initial calculation, we have 7 valid triangles out of 10 combinations. Comparing this with the options given, the correct choice is \(^5C_3 - 3 = 7\).

Key Concepts

Combinatorial Selection

Combinatorial selection is a mathematical technique used to count distinct groups or selections from a larger set. In this concept, we use combinations to determine how many ways we can choose a subset of items from a larger pool, without considering the order of selection.

To determine the number of triangles we can form from five line segments, we calculate the total combinations of three segments from the five given lengths. This is done using the combination formula, represented as \[ ^nC_k = \frac{n!}{k!(n-k)!} \]where \( n \) is the total number of items and \( k \) is the number of items to choose.

For our specific problem, we have 5 line segments and need to select 3, so we compute \[ ^5C_3 = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10 \] This calculation tells us that there are 10 possible ways to select 3 segments from the 5 available.

Geometry Problems

Geometry problems often involve understanding shapes, sizes, and the properties of space. Here, we are dealing with triangles, which require an understanding of several principles, especially the Triangle Inequality Theorem.

This theorem is fundamental when solving problems about forming triangles from given lines. It asserts that for any triangle with sides \( a \), \( b \), and \( c \), the sum of any two sides must be greater than the third side. Specifically:

• \(a + b > c\)
• \(a + c > b\)
• \(b + c > a\)

For instance, using this theorem, we assess each set of three line segments to verify if they can form a triangle. In our problem, out of the 10 possible combinations, only those that satisfy the triangle inequality theorem are valid triangles.

Triangles from Line Segments

Creating a triangle from line segments involves ensuring the three chosen segments can physically form a closed shape. Through the exercise, we learn how to apply the Triangle Inequality Theorem practically.

Given line segments of lengths 2, 3, 4, 5, and 6, we work through each triplet of segments to check which ones meet the conditions for forming a triangle. By systematically checking all combinations, we found that there are 7 combinations that adhere to the theorem: (2, 3, 4), (2, 4, 5), (2, 5, 6), (3, 4, 5), (3, 4, 6), (3, 5, 6), and (4, 5, 6).
Understanding how these segments combine helps us see the geometric and combinatorial principles at play, ultimately reinforcing our comprehension of both geometry and simple combinatorial logic.