JEE Main 2024 (06 Apr Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Permutations and Combinations: The number of triangles whose vertices are at the vertices of a regular octagon but none of whose sides is a side of the octagon is

Q: Why are there 4 invalid vertices for the 3rd point when picking 1 side?

Let's visualize it. If we pick a side, say $V_1V_2$, the third vertex cannot be $V_1$ or $V_2$ (otherwise it's not a triangle). It also cannot be $V_8$ or $V_3$ because if we picked those, the triangle would share two sides with the octagon ($V_8V_1$ and $V_1V_2$, or $V_1V_2$ and $V_2V_3$). So, we exclude the 2 endpoints and the 2 immediate neighbors, leaving $8 - 4 = 4$ valid choices.

Q: Is there a general formula for this?

Yes! For a polygon with $n$ vertices, the number of triangles with no sides in common with the polygon is $\frac{n(n-4)(n-5)}{6}$. However, in the JEE, memorizing formulas is risky. Understanding the 'Total - Unwanted' logic is much more powerful because it works even if the problem adds extra constraints!

Select Answer:

Progressive Hints

Hint 1

Hint 1: Imagine the octagon as a circle with 8 points. Instead of trying to draw every possible triangle, think about the total number of ways to pick any 3 vertices, and then subtract the cases that violate the rule (i.e., triangles that use 1 or 2 sides of the octagon).

Hint 2

Hint 2: To count triangles with 2 sides in common, notice that such a triangle is formed by 3 consecutive vertices. How many sets of 3 consecutive vertices exist in an octagon? For triangles with 1 side in common, fix one side of the octagon first, then choose the third vertex such that it is not an endpoint of that side and not an adjacent vertex.

Hint 3

Hint 3: The total number of triangles is given by

(3 8)

. For 2-side common triangles, there are 8 such triangles. For 1-side common triangles, for each of the 8 sides, you have

8 - 4 = 4

choices for the third vertex. The final answer is

(3 8) - (8 + 8 \times 4)

Visualized Solution

The Regular Octagon

A regular octagon has $n = 8$ vertices.
We need to form triangles using these vertices.
Condition: No side of the triangle can be a side of the octagon.

Complementary Counting Strategy

Direct counting is complex, so we use the complementary method.
$Required = Total - (1 side common + 2 sides common)$

Total Possible Triangles

To form any triangle, we select 3 vertices out of 8.
$Total Triangles =^{8} C_{3}$

Calculating Total Triangles

$^{8} C_{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1}$
$Total Triangles = 56$

Triangles with Two Common Sides

Let's visualize triangles that share exactly two sides with the octagon.
These are formed by taking 3 consecutive vertices (e.g., $V_{1}, V_{2}, V_{3}$ ).

Counting Two-Side Common Triangles

Each such triangle corresponds to a unique vertex in the middle.
For $n = 8$ , there are exactly $8$ such triangles.

Triangles with One Common Side

Now, let's visualize triangles sharing exactly one side with the octagon.
First, pick 1 side of the octagon ( $8$ choices).

Selecting the Third Vertex

The 3rd vertex cannot be the endpoints of the chosen side.
It also cannot be the adjacent vertices, otherwise it forms a 2-side common triangle.

Counting One-Side Common Triangles

Invalid vertices for 3rd point: $4$ (2 endpoints + 2 neighbors).
Valid choices for 3rd vertex: $8 - 4 = 4$ .
Total such triangles: $8 \times 4 = 32$ .

Assembling the Equation

Substitute the values back into our complementary counting equation:
$Required = 56 - (32 + 8)$

Final Calculation

$Required = 56 - 40$
$Required = 16$

Visualizing the Valid Triangles

The $16$ triangles are formed entirely by the diagonals of the octagon.
General Formula: $\frac{n \cdot ^{n - 4} C _{2}}{3}$

00:00 / 00:00

The Sigma Insight: Combinations and Selection

Solution Diagram

Analyzing the Setup

To determine the number of triangles that can be formed using the vertices of a regular octagon such that no side of the triangle is a side of the octagon, we employ the principle of complementary counting. We define the "Total Universe" as all possible triangles formed by any 3 vertices of the octagon.

The total number of ways to choose 3 vertices out of 8 is given by the combination formula:

Total Triangles = (3 8) = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56

The Unwanted Guests

Two-Side Common

Next, we identify triangles that share exactly two sides with the octagon. These triangles are formed by selecting three consecutive vertices (e.g.,

V_{1}, V_{2}, V_{3}

Since there are 8 vertices, there are exactly 8 such triplets of consecutive vertices. Each triplet forms a triangle that uses two sides of the octagon.

2-Side Common = 8

The Unwanted Guests

One-Side Common

Now, we count triangles that share exactly one side with the octagon. We first select one of the 8 sides of the octagon.

For a chosen side, say

V_{1} V_{2}

, we must select a third vertex. We must exclude

V_{1}

and

V_{2}

to ensure a triangle is formed, and we must also exclude the two adjacent vertices (

V_{8}

and

V_{3}

) to avoid counting triangles that share two sides.

This leaves

8 - 4 = 4

possible vertices for each of the 8 sides. Therefore:

1-Side Common = 8 \times 4 = 32

Final Calculation

To find the number of triangles that share no sides with the octagon, we subtract the "unwanted" triangles from the total number of possible triangles.

Required Triangles = Total - (1-Side Common + 2-Side Common)

Substituting our calculated values:

Required Triangles = 56 - (32 + 8) = 56 - 40 = 16

The total number of valid triangles is 16.

Common Traps & FAQs

Wait, why do we subtract the 2-side common triangles separately? Can't we just count 1-side common ones?

That is a great question! In combinatorics, we must ensure our sets are mutually exclusive. A triangle with 2 sides in common is a distinct geometric entity from a triangle with only 1 side in common. If we only subtracted the 1-side common triangles, we would still have those 2-side common triangles lurking in our total, leading to an incorrect answer. We must remove all 'unwanted' configurations completely.

Why are there 4 invalid vertices for the 3rd point when picking 1 side?

Let's visualize it. If we pick a side, say

V_{1} V_{2}

, the third vertex cannot be

V_{1}

V_{2}

(otherwise it's not a triangle). It also cannot be

V_{8}

V_{3}

because if we picked those, the triangle would share two sides with the octagon (

V_{8} V_{1}

and

V_{1} V_{2}

, or

V_{1} V_{2}

and

V_{2} V_{3}

). So, we exclude the 2 endpoints and the 2 immediate neighbors, leaving

8 - 4 = 4

valid choices.

Is there a general formula for this?

Yes! For a polygon with

n

vertices, the number of triangles with no sides in common with the polygon is

\frac{n ( n - 4 ) ( n - 5 )}{6}

. However, in the JEE, memorizing formulas is risky. Understanding the 'Total - Unwanted' logic is much more powerful because it works even if the problem adds extra constraints!

Animated Solution for Mathematics - Permutations and Combinations: The number of triangles whose vertices are at the vertices of a regular octagon but none of whose sides is a side of the octagon is

Select Answer:

Visualized Solution

The Regular Octagon

Complementary Counting Strategy

Total Possible Triangles

Calculating Total Triangles

Triangles with Two Common Sides

Counting Two-Side Common Triangles

Triangles with One Common Side

Selecting the Third Vertex

Counting One-Side Common Triangles

Assembling the Equation

Final Calculation

Visualizing the Valid Triangles

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Unwanted Guests

The Unwanted Guests

Final Calculation

Similar Questions

There are 12 points in a plane, no three of which are in the same straight line, except 5 points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these 12 points is

The side $A B, B C$ and $C A$ of a triangle $A B C$ have 3, 4 and 5 interior points respectively on them. The number of triangles that can be constructed using these interior points as vertices is .........

If the sides AB, BC and CA of a triangle ABC have 3,5 and 6 interior points respectively, then the total number of triangles that can be constructed using these points as vertices, is equal to :

Let $T_{n}$ be the number of all possible triangles formed by joining vertices of an $n$ -sided regular polygon. If $T_{n + 1} - T_{n} = 10$ , then the value of $n$ is :

Let $P_{1}, P_{2}, \dots, P_{15}$ be 15 points on a circle. The number of distinct triangles formed by points $P_{i}, P_{j}, P_{k}$ such that $i + j + k \neq = 15$ , is :

Let $T_{n}$ denote the number of triangles which can be formed using the vertices of a regular polygon of $n$ sides. If $T_{n + 1} - T_{n} = 21$ , then $n$ equals

Animated Solution for Mathematics - Permutations and Combinations: The number of triangles whose vertices are at the vertices of a regular octagon but none of whose sides is a side of the octagon is

Select Answer:

Visualized Solution

The Regular Octagon

Complementary Counting Strategy

Total Possible Triangles

Calculating Total Triangles

Triangles with Two Common Sides

Counting Two-Side Common Triangles

Triangles with One Common Side

Selecting the Third Vertex

Counting One-Side Common Triangles

Assembling the Equation

Final Calculation

Visualizing the Valid Triangles

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Unwanted Guests

The Unwanted Guests

Final Calculation

Similar Questions

There are 12 points in a plane, no three of which are in the same straight line, except 5 points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these 12 points is

If the sides AB, BC and CA of a triangle ABC have 3,5 and 6 interior points respectively, then the total number of triangles that can be constructed using these points as vertices, is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Tn​ be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If Tn+1​−Tn​=10, then the value of n is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P1​,P2​,…,P15​ be 15 points on a circle. The number of distinct triangles formed by points Pi​,Pj​,Pk​ such that i+j+k=15, is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Tn​ denote the number of triangles which can be formed using the vertices of a regular polygon of n sides. If Tn+1​−Tn​=21, then n equals

Let $T_{n}$ be the number of all possible triangles formed by joining vertices of an $n$ -sided regular polygon. If $T_{n + 1} - T_{n} = 10$ , then the value of $n$ is :

Let $P_{1}, P_{2}, \dots, P_{15}$ be 15 points on a circle. The number of distinct triangles formed by points $P_{i}, P_{j}, P_{k}$ such that $i + j + k \neq = 15$ , is :

Let $T_{n}$ denote the number of triangles which can be formed using the vertices of a regular polygon of $n$ sides. If $T_{n + 1} - T_{n} = 21$ , then $n$ equals