Combinatorics

To find the number of triangles with vertices on different sides of the convex quadrilateral PQRS, we need to consider points on each side of the quadrilateral. Analyzing what is given:

3 points on side PQ
4 points on side QR
5 points on side RS
6 points on side PS

To form a triangle, we need to select one point from each of the three different sides of the quadrilateral. Therefore, we need to compute the possible combinations of selecting points from any trio of these sides.

The sides can be categorized into groups of three in the following combinations:

Select one point from side PQ, one point from side QR, and one point from side RS. $C(3%2C%201)%20%5Ctimes%20C(4%2C%201)%20%5Ctimes%20C(5%2C%201)%20%3D%203%20%5Ctimes%204%20%5Ctimes%205%20%3D%2060$
Select one point from side PQ, one point from side RS, and one point from side PS. $C(3%2C%201)%20%5Ctimes%20C(5%2C%201)%20%5Ctimes%20C(6%2C%201)%20%3D%203%20%5Ctimes%205%20%5Ctimes%206%20%3D%2090$
Select one point from side PQ, one point from side PS, and one point from side QR. $C(3%2C%201)%20%5Ctimes%20C(4%2C%201)%20%5Ctimes%20C(6%2C%201)%20%3D%203%20%5Ctimes%204%20%5Ctimes%206%20%3D%2072$
Select one point from side QR, one point from side RS, and one point from side PS. $C(4%2C%201)%20%5Ctimes%20C(5%2C%201)%20%5Ctimes%20C(6%2C%201)%20%3D%204%20%5Ctimes%205%20%5Ctimes%206%20%3D%20120$

Adding all these computations together gives the total number of triangles possible:

$60%20%2B%2090%20%2B%2072%20%2B%20120%20%3D%20342$

Thus, the correct answer is 342.

Step 1: Understanding the Concept:
Use the property that

%7B%7D%5EnC_r%20%3D%20%7B%7D%5EnC_%7Bn-r%7D

to relate

S_n

and

t_n

.

Step 2: Detailed Explanation:
Write

S_n%20%3D%20%5Csum_%7Br%3D0%7D%5En%20%5Cfrac%7Br%7D%7B(%7B%7D%5EnC_r)%5Ek%7D

. Replace

r

n-r

S_n%20%3D%20%5Csum_%7Br%3D0%7D%5En%20%5Cfrac%7Bn-r%7D%7B(%7B%7D%5EnC_%7Bn-r%7D)%5Ek%7D%20%3D%20%5Csum_%7Br%3D0%7D%5En%20%5Cfrac%7Bn-r%7D%7B(%7B%7D%5EnC_r)%5Ek%7D

Adding the two expressions for

S_n

2S_n%20%3D%20%5Csum_%7Br%3D0%7D%5En%20%5Cfrac%7Br%20%2B%20n%20-%20r%7D%7B(%7B%7D%5EnC_r)%5Ek%7D%20%3D%20n%20%5Csum_%7Br%3D0%7D%5En%20%5Cfrac%7B1%7D%7B(%7B%7D%5EnC_r)%5Ek%7D%20%3D%20n%20t_n

Thus

%5Cfrac%7BS_n%7D%7Bn%20t_n%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D

. Then

%5Ccos%5E%7B-1%7D%5Cleft(%5Cfrac%7B1%7D%7B2%7D%5Cright)%20%3D%20%5Cfrac%7B%5Cpi%7D%7B3%7D

. Wait, there is a mis-step: The sum

S_n

is defined with

r

in numerator. The property

%7B%7D%5EnC_r%20%3D%20%7B%7D%5EnC_%7Bn-r%7D

gives:

S_n%20%3D%20%5Csum_%7Br%3D0%7D%5En%20%5Cfrac%7Br%7D%7B(%7B%7D%5EnC_r)%5Ek%7D%20%3D%20%5Csum_%7Br%3D0%7D%5En%20%5Cfrac%7Bn-r%7D%7B(%7B%7D%5EnC_%7Bn-r%7D)%5Ek%7D%20%3D%20%5Csum_%7Br%3D0%7D%5En%20%5Cfrac%7Bn-r%7D%7B(%7B%7D%5EnC_r)%5Ek%7D

Adding:

2S_n%20%3D%20%5Csum_%7Br%3D0%7D%5En%20%5Cfrac%7Bn%7D%7B(%7B%7D%5EnC_r)%5Ek%7D%20%3D%20n%20t_n

. So

%5Cfrac%7BS_n%7D%7Bn%20t_n%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D

. Thus

%5Ccos%5E%7B-1%7D(1%2F2)%20%3D%20%5Cpi%2F3

. But the answer given in many sources is

%5Cpi%2F2

. Let's re-check: If

S_n%20%3D%20%5Csum%20r%2F(C_r)%5Ek

and

t_n%20%3D%20%5Csum%201%2F(C_r)%5Ek

, then

S_n%2Ft_n

is the average of

r

, which for symmetric binomial coefficients is

n%2F2

. So

S_n%2F(n%20t_n)%20%3D%201%2F2

. Then

%5Ccos%5E%7B-1%7D(1%2F2)%20%3D%20%5Cpi%2F3

. However, the problem likely intends

%5Ccos%5E%7B-1%7D(S_n%2F(n%20t_n))%20%3D%20%5Cpi%2F3

. But the given answer choices have

%5Cpi%2F2

as well. Re-evaluate: There is a known identity:

%5Csum%20r%2F%7B%7D%5EnC_r%20%3D%20n%2F2%20%5Csum%201%2F%7B%7D%5EnC_r

. So ratio is

n%2F2

. Then

%5Ccos%5E%7B-1%7D((n%2F2)%2F(n))%20%3D%20%5Ccos%5E%7B-1%7D(1%2F2)%20%3D%20%5Cpi%2F3

. But if the denominator is

n%20t_n

, then it's

%5Ccos%5E%7B-1%7D((n%20t_n%2F2)%2F(n%20t_n))%20%3D%20%5Ccos%5E%7B-1%7D(1%2F2)%20%3D%20%5Cpi%2F3

. So answer should be

%5Cpi%2F3

. But the provided answer key says

%5Cpi%2F2

. Possibly a misprint. Following the derivation:

S_n%20%3D%20%5Csum%20r%20a_r

t_n%20%3D%20%5Csum%20a_r

with

a_r%20%3D%201%2F(%7B%7D%5EnC_r)%5Ek

. Since

a_r%20%3D%20a_%7Bn-r%7D

S_n%20%3D%20%5Csum%20(n-r)a_r%20%3D%20n%20t_n%20-%20S_n%20%5CRightarrow%202S_n%20%3D%20n%20t_n%20%5CRightarrow%20S_n%2F(n%20t_n)%20%3D%201%2F2

. So

%5Ccos%5E%7B-1%7D(1%2F2)%20%3D%20%5Cpi%2F3

. I will present the derived answer.

Step 3: Final Answer:

%5Cfrac%7BS_n%7D%7Bn%20t_n%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D

, so

%5Ccos%5E%7B-1%7D%5Cleft(%5Cfrac%7B1%7D%7B2%7D%5Cright)%20%3D%20%5Cfrac%7B%5Cpi%7D%7B3%7D

, which corresponds to option (B).

About Combinatorics - MET

Combinatorics is a vital chapter for MET aspirants. Mastering the concepts covered in this chapter is essential for securing a top rank.

By rigorously practicing the previous year questions associated with this chapter, you can identify high-yield topics, understand the examiner's perspective, and boost your confidence during the actual exam.

Frequently Asked Questions

Why focus on Combinatorics PYQs?

Analyzing PYQs for this specific chapter reveals the most frequently tested concepts and the typical complexity of questions, allowing you to tailor your study plan efficiently.

How to best use this analysis?

Review the topic breakdown to see which sub-topics within Combinatorics carry the most weight. Then, tackle the questions iteratively to solidify your understanding.

High-Yield Trend

Chapter Questions
3 MCQs

Official Solution

Official Solution

Official Solution

About Combinatorics - MET

Frequently Asked Questions

Why focus on Combinatorics PYQs?

How to best use this analysis?

Combinatorics

High-Yield Trend

Chapter Questions 3 MCQs

Official Solution

Official Solution

Official Solution

About Combinatorics - MET

Frequently Asked Questions

Why focus on Combinatorics PYQs?

How to best use this analysis?

Chapter Questions
3 MCQs