Permutations

Permutations

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Permutation is the ordered arrangement of objects selected from a set. For n distinct objects taken r at a time, the count is nPr = n! / (n − r)!, with 0! = 1 and P(n, n) = n!.

• When r positions each allow any of n objects (repetition allowed), total arrangements = n^r.
• Circular arrangements of n distinct objects = (n − 1)!, since rotations collapse.
• Permutations of n objects with repeats p₁, p₂, …, pₖ = n! / (p₁! · p₂! · … · pₖ!).
• Always define variables: n = total pool, r = positions to fill, r ≤ n, unitless.
• CUET trap: swapping nPr with nCr loses the r! factor and produces wrong answers in word-formation MCQs.

Term	Meaning
nPr	Ordered selection of r from n
n!	Product 1·2·…·n
Circular	(n−1)! arrangements

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Core Definition and Formula

A permutation of r objects chosen from n distinct objects is an ordered arrangement where every sequence is distinguishable. The notation nPr (also written P(n, r)) counts such arrangements:

nPr = n! / (n − r)!, where n ≥ r ≥ 0 are integers.

Special cases: P(n, n) = n! (arrange all), and P(n, 0) = 1 (empty arrangement, defined for the Fundamental Principle of Counting). The factorial grows explosively — 10! = 3,628,800 — so CUET problems in most keys cap n at 8 or below.

Worked Relationships

The Fundamental Principle of Counting underpins the formula: filling position 1 has n choices, position 2 has n − 1, and so on down to position r, giving the product n · (n−1) · … · (n−r+1) = n!/(n−r)!.

Key linkage: nPr = nCr × r!, which lets you convert between ordered and unordered selections. For example, choosing a President, Vice-President, and Secretary from 12 students gives P(12, 3) = 12 · 11 · 10 = 1320 ordered outcomes.

Concept	Formula / Value	When to use
Linear, all distinct	n!	Arrange every object
Linear, r of n	n! / (n−r)!	Pick subset, order matters
Circular, distinct	(n−1)!	Seating around a table
With repetition (r slots, n choices)	n^r	Digits, codes with repeats
With identical objects	n! / (p₁!·…·pₖ!)	Words like MISSISSIPPI

Typical CUET Pattern

Expect 1–2 MCQs of 1 mark each, in most keys a numeric computation (e.g., “How many 4-digit codes can be formed using digits 0–9 with no repetition?”). The trap answer is 9 × 9 × 8 × 7 (forgetting the leading-zero restriction), while the correct path uses 9 · P(9, 3) = 9 · 504 = 4536.

• Restriction strategy: place the most constrained object first, then arrange the rest freely.
• Always check whether clockwise and counterclockwise rotations are counted as distinct — most CUET seating problems treat them as identical.
• Verify the r ≤ n condition before applying nPr; otherwise the problem is actually about repetition.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Edge Cases and Derangements

When no object can occupy its original position, the count is a derangement D(n):

D(n) = n! · (1 − 1/1! + 1/2! − 1/3! + … + (−1)ⁿ / n!).

For n = 4, D(4) = 24 · (1 − 1 + 0.5 − 0.1667 + 0.0417) = 24 · 0.375 = 9. CUET rarely tests this formula directly but it appears disguised as “in how many ways can 5 letters be placed in 5 envelopes so that none goes to the right address?”

Adjacency and Gap Methods

Two flavours of restricted arrangement dominate exams:

1. Bundle method: treat adjacent objects as a single block (e.g., “A and B must sit together” → arrange (n−1)! · 2! blocks).
2. Gap method: arrange the unconstrained objects first, then slot the constrained ones into the gaps between them — useful when specific objects must not be adjacent.

For 8 people around a circular table where two particular people must sit together, the count is 2 · 6! = 1440 (bundle of 2 × internal arrangements × circular permutations of 7).

Common Mistakes

Mistake	Correction
Using nCr where order matters	Multiply nCr by r!
Counting n! for a circle	Subtract factor n → (n−1)!
Ignoring identical letters in words	Divide by pᵢ! for each repeat
Sign error in derangement sum	Alternate +/− strictly from term 2

Practice Prompts

1. How many distinct arrangements of the letters of BANANA exist? (Answer: 6! / (3! · 2! · 1!) = 60.)
2. In how many ways can 6 boys and 4 girls be seated in a row if the girls sit together? (Answer: 7! · 4! = 120960.)

Exam Strategy for CUET UG

Permutations carries roughly 3% weight in the Mathematics domain and overlaps heavily with probability, binomial expansion (nCr identities), and determinant properties (cofactor expansion). Allocate 60–90 seconds per MCQ; skip and flag if n > 12 because factorial overflow is a sign of a misread question. The NTA official syllabus lists permutations under “Algebra — Counting Principles,” and NCERT Class 11 textbook (Chapter 7) is the most reliable source for derivation drills.

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