In-Depth: Permutations and Combinations

CLAT Application & Relevance

Importance: Very Low. Direct, complex questions involving Permutations and Combinations are highly unlikely in the modern CLAT QT section, which focuses on Data Interpretation. However, the fundamental concepts of counting (arrangement vs. selection) can occasionally surface in very simple logical reasoning or numerical caselets. It's best to understand the basic distinction and formulas without delving into advanced problem types.

How it's tested: Extremely rarely, as a basic counting problem embedded in a larger textual context, possibly asking for the number of ways to arrange simple items or select a small group.

Section 1: Core Concepts & Formulas

Permutations and Combinations are about counting the number of ways to arrange or select items from a set. The key distinction lies in whether the order of arrangement matters.

Key Definitions

Factorial (n!): The product of all positive integers less than or equal to n. n! = n × (n-1) × (n-2) × ... × 1. By definition, 0! = 1.
Permutation: An arrangement of items where the order matters. (e.g., arranging letters in a word, seating people in specific chairs).
Combination: A selection of items where the order does NOT matter. (e.g., choosing a committee, selecting cards for a hand).

Fundamental Formulas

Number of Permutations (nPr): The number of ways to arrange 'r' distinct items chosen from 'n' distinct items.
nPr = n! / (n-r)!
Number of Combinations (nCr): The number of ways to select 'r' distinct items from 'n' distinct items.
nCr = n! / (r! * (n-r)!)
Note: nCr = nPr / r!
Permutations of items with repetition: If there are 'n' items where p1 are of one type, p2 of another, etc., then total distinct permutations = n! / (p1! * p2! * ...)

Section 2: Solved CLAT-Style Examples (Basic Application)

Example 1: Basic Permutation (Arrangement)

Passage Context: "A legal panel consists of 5 members. They are to be seated in a row of 5 chairs for a press conference. The arrangement matters for seating protocol."

Question A: "In how many different ways can the 5 members be seated in the 5 chairs?"

Question B: "If only 3 out of the 5 members are to be selected and arranged in 3 specific chairs, in how many ways can this be done?"

Detailed Solution A:
1. Identify as Permutation: Order matters (seating arrangement).
2. Number of items (n) = 5, Number to arrange (r) = 5. This is 5P5 or simply 5!.
3. Calculate: 5! = 5 × 4 × 3 × 2 × 1 = 120.
Answer A: The 5 members can be seated in 120 different ways.

Detailed Solution B:
1. Identify as Permutation: Order matters (arranging in specific chairs).
2. Number of items (n) = 5, Number to arrange (r) = 3. This is 5P3.
3. Apply Formula: nPr = n! / (n-r)!
5P3 = 5! / (5-3)! = 5! / 2!
= (5 × 4 × 3 × 2 × 1) / (2 × 1)
= 5 × 4 × 3 = 60.
Answer B: This can be done in 60 ways.

Example 2: Basic Combination (Selection)

Passage Context: "A jury pool consists of 10 qualified individuals. From this pool, a 3-member special committee needs to be formed to review complex evidence. The order in which members are chosen for the committee does not matter, only who is on it."

Question: "In how many different ways can this 3-member committee be selected?"

Detailed Solution:
1. Identify as Combination: Order does NOT matter (forming a committee).
2. Number of items (n) = 10, Number to select (r) = 3. This is 10C3.
3. Apply Formula: nCr = n! / (r! * (n-r)!)
10C3 = 10! / (3! * (10-3)!) = 10! / (3! * 7!)
= (10 × 9 × 8 × 7!) / ((3 × 2 × 1) * 7!)
= (10 × 9 × 8) / (3 × 2 × 1)
= 10 × 3 × 4 = 120.
Answer: The 3-member committee can be selected in 120 different ways.