Permutations and Combinations – JEE Main Maths Formula Sheet & Class 11 Notes | All Formulas, Results & Applications

Fundamental Principle of Counting — Multiplication and Addition Theorems

Why the Counting Principles Are the Foundation of All Permutations and Combinations Formulas

All Permutations and Combinations formulas ultimately rest on two counting principles that determine when to multiply and when to add.

Multiplication Theorem (AND rule) in Permutations and Combinations: If one operation can be performed in m ways, AND another independent operation can subsequently be performed in n ways, then BOTH operations together can be performed in m × n ways. This is used when successive operations must ALL be completed — multiply the choices. Extension: if k sequential independent operations have n₁, n₂, …, nₖ ways respectively, the total number of ways = n₁ × n₂ × … × nₖ. Example: A lock has 4 digits each from 0–9, total codes = 10 × 10 × 10 × 10 = 10⁴.

Addition Theorem (OR rule) in Permutations and Combinations: If one operation can be performed in m ways, OR another mutually exclusive operation can be performed in n ways, then EITHER operation can be performed in m + n ways. This is used when alternatives are being counted — add the choices. Example: if a student can choose a book from 3 science books OR 4 history books, total choices = 3 + 4 = 7. Key distinction for Permutations and Combinations JEE Main: AND → multiply; OR/EITHER → add. Most problems combine both principles in sequence.

Factorial notation in Permutations and Combinations: n! (n factorial) = n × (n–1) × (n–2) × … × 3 × 2 × 1. Recursive: n! = n × (n–1)!. Special values: 0! = 1 (by convention — critical for Permutations and Combinations formulas), 1! = 1. n! is defined only for non-negative integers; (–r)! is undefined. The formula ⁿPᵣ = n!/(n–r)! requires (n–r)! to be defined, so r ≤ n. Download the Free PDF for Permutations and Combinations for all factorial examples and counting principle applications.

Permutations Formula ⁿPᵣ — All Standard Cases and Restrictions

Why All Permutation Restriction Formulas Are Direct JEE Main One-Mark Questions

Core permutation formula in Permutations and Combinations: ⁿPᵣ = n!/(n–r)! = n(n–1)(n–2)…(n–r+1). This is the number of ways to arrange r objects chosen from n distinct objects where order matters. ⁿPₙ = n! (arrange all n). ⁿP₁ = n. ⁿP₀ = 1. ⁿP₂ = n(n–1). ⁿPᵣ = r! · ⁿCᵣ.

Permutation with identical items (Permutations and Combinations formula): If n things are taken all at a time with p alike of one kind, q alike of second kind, and rest all different: number of permutations = n! / (p! · q!). Extension: with p₁ alike of type 1, p₂ alike of type 2, …, pₖ alike of type k: number = n! / (p₁! · p₂! · … · pₖ!) where p₁+p₂+…+pₖ ≤ n. Example: number of arrangements of MISSISSIPPI = 11!/(4!4!2!1!) (1 M, 4 I, 4 S, 2 P).

Permutation with repetition allowed (Permutations and Combinations): Number of permutations of n different things taken r at a time when repetition is allowed = nʳ. Example: number of 4-digit PIN codes from digits 0–9 with repetition = 10⁴.

Permutation restriction formulas in Permutations and Combinations (JEE Main direct questions):

1. When one particular item is ALWAYS included in r-arrangement from n: r · ⁿ⁻¹Pᵣ₋₁. (Choose position for the fixed item in r ways, then arrange remaining r–1 from n–1.)

2. When p particular items are ALWAYS included: ⁿ⁻ᵖPᵣ₋ₚ · r!/(r–p)! = Pᵣ₋ₚ/(r–p)! · r! → simplification: ⁿ⁻ᵖPᵣ₋ₚ · ⁿP₀ · r!/(r–p)! → final = ⁿ⁻ᵖPᵣ₋ₚ · Pₚ/... → the direct formula from PDF: [r!/(r–p)!] · ⁿ⁻ᵖPᵣ₋ₚ.

3. When p particular items are ALWAYS TOGETHER (treated as one block): p! · (n–p+1)! · ⁿ⁻ᵖ⁺¹Pᵣ₋ₚ. For r = n (all items): p! · (n–p+1)!.

4. When p particular items are NEVER included: ⁿ⁻ᵖPᵣ.

5. When p particular items always OCCUR TOGETHER (all n items): p! · (n–p+1)!.

6. When m particular items NEVER occur together (all n items): n! – [m! · (n–m+1)!].

7. When r particular items are in a SPECIFIC FIXED ORDER (all n arranged): n!/r!. (Divide by r! because only 1 of r! orders of the r fixed items is allowed.)

Circular Permutations — Formulas for JEE Main

Why Circular Permutation Formulas Are High-Frequency JEE Main Permutations and Combinations Topics

Circular permutations in Permutations and Combinations differ from linear permutations because one position is fixed (the "reference" position) to break the rotational symmetry — this removes one degree of freedom.

Circular permutation of n distinct things (around a round table): When clockwise and anticlockwise arrangements are considered DIFFERENT: (n–1)!. This is because in a circle, we fix one person and arrange the remaining (n–1) in (n–1)! ways. Number of linear permutations = n!; dividing by n (for n equivalent rotations) gives (n–1)!.

Circular permutation (necklace/garland) when clockwise = anticlockwise: When both orientations are the SAME (necklace can be flipped): (n–1)!/2. This is because each arrangement and its mirror image are counted as one. Used for necklaces, garlands, bracelets.

Circular permutation of n things taken r at a time: When clockwise ≠ anticlockwise: ⁿPᵣ/r. When clockwise = anticlockwise: ⁿPᵣ/(2r).

Key circular permutation results in Permutations and Combinations: Circular arrangement of n persons where 2 specific persons are ALWAYS together = 2! · (n–2)! = 2(n–2)! (treat two as one block → (n–1) in circle = (n–2)!, then internally 2! arrangements). Circular arrangement where 2 specific persons are NEVER together = (n–1)! – 2(n–2)! = (n–2)!(n–3). Number of necklaces with n beads of different colours = (n–1)!/2. Download the Free PDF for Permutations and Combinations for all circular permutation worked examples.

Combinations Formula ⁿCᵣ — All Standard Results and Identities for JEE Main

Why Combination Identities Are the Most Formula-Dense Section of Permutations and Combinations

Core combination formula in Permutations and Combinations: ⁿCᵣ = n! / (r! · (n–r)!) = ⁿPᵣ / r! . This counts the number of ways to SELECT r objects from n distinct objects where order does NOT matter. Defined for 0 ≤ r ≤ n; ⁿCᵣ = 0 if r > n.

Standard combination identities in Permutations and Combinations (JEE Main formulas — all must be memorised):

(i) ⁿCᵣ = ⁿPᵣ / r! = n! / (r!(n–r)!)

(ii) ⁿC₀ = ⁿCₙ = 1

(iii) ⁿC₁ = ⁿCₙ₋₁ = n

(iv) ⁿCᵣ = ⁿCₙ₋ᵣ (symmetry property — most used combination identity in JEE Main)

(v) ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ (Pascal's Rule / Pascal's Triangle identity)

(vi) n · ⁿ⁻¹Cᵣ₋₁ = (n–r+1) · ⁿCᵣ₋₁ = r · ⁿCᵣ → ⁿCᵣ/ⁿCᵣ₋₁ = (n–r+1)/r

(vii) ⁿCᵣ₋₁/ⁿCᵣ = r/(n–r+1)

(viii) Maximum value of ⁿCᵣ: occurs at r = n/2 if n is even (ⁿCₙ/₂ is maximum); at r = (n–1)/2 or r = (n+1)/2 if n is odd (both equal — two equal maxima).

(ix) Sum of all combinations: ⁿC₀ + ⁿC₁ + ⁿC₂ + … + ⁿCₙ = 2ⁿ (put x=1 in binomial theorem (1+x)ⁿ)

(x) Sum of alternate terms: ⁿC₀ + ⁿC₂ + ⁿC₄ + … = 2ⁿ⁻¹ (even-position). ⁿC₁ + ⁿC₃ + ⁿC₅ + … = 2ⁿ⁻¹ (odd-position).

(xi) ⁿC₁ + ⁿC₂ + … + ⁿCₙ = 2ⁿ – 1 (at least one selection from n things)

(xii) ⁿC₀ – ⁿC₁ + ⁿC₂ – ⁿC₃ + … + (–1)ⁿ·ⁿCₙ = 0 (put x = –1 in binomial theorem)

(xiii) Sum of even-indexed = sum of odd-indexed = 2ⁿ⁻¹

(xiv) ⁿC₀² + ⁿC₁² + ⁿC₂² + … + ⁿCₙ² = ²ⁿCₙ (Vandermonde's identity special case)

(xv) Σᵣ₌₀ⁿ r·ⁿCᵣ = n·2ⁿ⁻¹

(xvi) Σᵣ₌₀ⁿ r²·ⁿCᵣ = n(n+1)·2ⁿ⁻² [use differentiation on (1+x)ⁿ]

Combination cases with restrictions (Permutations and Combinations JEE Main):

(a) Select r from n, k specific items ALWAYS included: ⁿ⁻ᵏCᵣ₋ₖ (already have k, choose r–k from remaining n–k).

(b) Select r from n, k specific items NEVER included: ⁿ⁻ᵏCᵣ (exclude k items, choose r from n–k).

(c) All possible selections from n distinct items (including empty set): ⁿC₀+ⁿC₁+…+ⁿCₙ = 2ⁿ.

(d) At least one item selected from n: 2ⁿ–1.

(e) Select r from n identical items: exactly 1 way (or 0 ways if r > n).

(f) Select none, one, or more from n identical items: (n+1) ways (0,1,2,...,n items).

(g) From (p+q+r+s) items where p alike type 1, q alike type 2, r alike type 3, s all different: total selections (including empty) = (p+1)(q+1)(r+1)·2ˢ; at least one = (p+1)(q+1)(r+1)·2ˢ – 1. Download the Free PDF for Permutations and Combinations for all combination identity proofs and applications.

Number of Divisors — Formula and Applications in Permutations and Combinations

Why Divisor Counting Formulas Are Direct JEE Main Permutations and Combinations Application Questions

Divisor counting uses the Permutations and Combinations framework applied to prime factorisation. These formulas appear as standalone JEE Main questions in the Permutations and Combinations section.

Let N = p₁^α₁ · p₂^α₂ · p₃^α₃ · … · pₖ^αₖ where p₁, p₂, …, pₖ are distinct primes and α₁, α₂, …, αₖ are positive integers. Then:

Total number of divisors of N (including 1 and N) = (α₁+1)(α₂+1)(α₃+1)…(αₖ+1). Each divisor is of the form p₁^b₁ · p₂^b₂ · … · pₖ^bₖ where 0 ≤ bᵢ ≤ αᵢ — giving (αᵢ+1) choices for each prime, multiplied over all primes.

Sum of all divisors of N (including 1 and N) = (p₁⁰ + p₁¹ + … + p₁^α₁)(p₂⁰ + p₂¹ + … + p₂^α₂)…(pₖ⁰ + pₖ¹ + … + pₖ^αₖ) = [(p₁^(α₁+1)–1)/(p₁–1)] · [(p₂^(α₂+1)–1)/(p₂–1)] · … · [(pₖ^(αₖ+1)–1)/(pₖ–1)].

Number of ways N resolves into product of two COPRIME factors = 2^(k–1) where k = number of distinct prime factors. (Each prime factor goes entirely into one of the two factors — 2 choices per prime, divided by 2 for order → 2ᵏ/2 = 2^(k–1); but if order matters = 2^(k–1) ordered pairs, if unordered = 2^(k–1) for unordered, 2ᵏ for ordered.)

Number of ways N resolves into product of two factors (unordered) = [(α₁+1)(α₂+1)…(αₖ+1)]/2 if N is NOT a perfect square; = [(α₁+1)(α₂+1)…(αₖ+1)+1]/2 if N IS a perfect square (the extra 1 accounts for the pair √N × √N where both factors are equal). Download the Free PDF for Permutations and Combinations for divisor formula worked examples.

Division into Groups and Distribution Problems in Permutations and Combinations

Why Distribution Formulas Are the Most Multi-Case Permutations and Combinations Section for JEE Main

Distribution problems in Permutations and Combinations ask: in how many ways can n objects be divided or distributed among groups? The answer depends critically on whether objects are distinct or identical, and whether groups are labelled (distinct) or unlabelled.

Division of (m+n) distinct objects into two groups of sizes m and n (m ≠ n) in Permutations and Combinations: Number of ways = C(m+n, m) = (m+n)! / (m! · n!). If m = n (equal groups) and groups are UNLABELLED (order doesn't matter): divide by 2! → (2n)! / (2 · (n!)²). If m = n and groups are LABELLED (distinct): (2n)! / (n!)².

Division of mn distinct objects equally into m groups of n each in Permutations and Combinations: If groups are UNLABELLED: (mn)! / ((n!)^m · m!). If groups are LABELLED (distinguishable): (mn)! / (n!)^m.

Distribution of n distinct objects into r distinct groups in Permutations and Combinations: Each of n objects can go into any of r groups → total (including empty groups) = rⁿ. If NO group is empty (inclusion-exclusion): = Σₖ₌₀ʳ (–1)ᵏ · ⁿCₖ · (r–k)ⁿ = rⁿ – ⁿC₁(r–1)ⁿ + ⁿC₂(r–2)ⁿ – … = Stirling numbers of second kind × r!.

Distribution of n identical objects into r distinct groups (Stars and Bars formula in Permutations and Combinations): Blank groups ALLOWED (each group can get 0 or more): ⁿ⁺ʳ⁻¹Cᵣ₋₁. Blank groups NOT allowed (each group must get at least 1): ⁿ⁻¹Cᵣ₋₁. These are the "stars and bars" formulas — also gives the number of non-negative integer solutions of x₁+x₂+…+xᵣ = n (= ⁿ⁺ʳ⁻¹Cᵣ₋₁) and positive integer solutions (= ⁿ⁻¹Cᵣ₋₁).

Distribution with minimum/maximum constraints in Permutations and Combinations: Number of ways to distribute n identical items in r groups so no group has fewer than l or more than m things = coefficient of xⁿ in (xˡ + xˡ⁺¹ + … + xᵐ)ʳ. Number of ways to select r things from n things where p are identical = ⁿ⁻ᵖCᵣ (if r ≤ p) + ⁿ⁻ᵖCᵣ₋₁ + … (use generating function). Download the Free PDF for Permutations and Combinations for all distribution formulas with examples.

Derangements Formula — Dₙ in Permutations and Combinations

Why the Derangement Formula Is a Unique JEE Main Permutations and Combinations Result

A derangement is a permutation of n objects such that NO object appears in its original position. The number of derangements of n objects, Dₙ, is a classic result in Permutations and Combinations derived using the inclusion-exclusion principle.

Derangement formula Dₙ in Permutations and Combinations:

Dₙ = n! · [1 – 1/1! + 1/2! – 1/3! + … + (–1)ⁿ/n!]

= n! · Σₖ₌₀ⁿ (–1)ᵏ/k!

Explicit values: D₁ = 0; D₂ = 1; D₃ = 2; D₄ = 9; D₅ = 44; D₆ = 265.

Recursive formula: Dₙ = (n–1)(Dₙ₋₁ + Dₙ₋₂) = n·Dₙ₋₁ + (–1)ⁿ.

As n→∞: Dₙ/n! → 1/e ≈ 0.368 (probability that a random permutation is a derangement ≈ 1/e).

Generalised derangement in Permutations and Combinations: If r things must go to WRONG places (out of n) and (n–r) go to CORRECT places: number of such arrangements = ⁿCₙ₋ᵣ · Dᵣ = ⁿCᵣ · Dᵣ. Classic JEE Main question type: in how many ways can 5 letters be placed in 5 envelopes so that exactly 2 go into correct envelopes? = ⁵C₂ · D₃ = 10 × 2 = 20. Download the Free PDF for Permutations and Combinations derangement formula derivation and applications.

Geometric Applications of Permutations and Combinations — Lines, Triangles, Circles, Rectangles

Why Geometric Counting Formulas Are the Most Application-Rich JEE Main Permutations and Combinations Section

Geometric counting problems in Permutations and Combinations apply ⁿCᵣ formulas to count geometric figures formed by points, lines, and circles. These appear regularly in JEE Main maths and can be answered directly using the formulas below.

Lines from points in Permutations and Combinations: Maximum number of lines from n points (no 3 collinear) = ⁿC₂ = n(n–1)/2. If m of the n points are collinear: actual number of lines = ⁿC₂ – ᵐC₂ + 1 (subtract lines among collinear points, add back the 1 line they all share).

Triangles from points in Permutations and Combinations: Maximum triangles from n points (no 3 collinear) = ⁿC₃. If m points are collinear: number of triangles = ⁿC₃ – ᵐC₃ (collinear points form 0 triangles; remove those combinations).

Diagonals of a polygon in Permutations and Combinations: Number of diagonals in a convex polygon of n sides = ⁿC₂ – n = n(n–1)/2 – n = n(n–3)/2. (Total lines from n vertices minus the n sides.)

Triangles in a polygon in Permutations and Combinations: Total triangles formed by vertices of n-gon = ⁿC₃. Triangles with NO side common with polygon = ⁿC₃ – n(n–4) – n. Triangles with exactly ONE side common = n(n–4). Triangles with exactly TWO sides common = n.

Intersection of lines and circles in Permutations and Combinations: Maximum intersection points of n distinct lines = ⁿC₂ (each pair intersects once). If m lines are concurrent: = ⁿC₂ – ᵐC₂ + 1. Maximum intersection points of n distinct circles = 2·ⁿC₂ = n(n–1) (each pair of circles intersects in at most 2 points). Maximum intersections of n circles AND m lines = 2·ⁿC₂ + ᵐC₂ + 2·ⁿC₁·ᵐC₁ = n(n–1) + m(m–1)/2 + 2mn.

Rectangles and squares in a grid (Permutations and Combinations): Number of rectangles (including squares) in n×p grid: choose 2 horizontal lines from (n+1) and 2 vertical lines from (p+1): = ⁿ⁺¹C₂ × ᵖ⁺¹C₂ = [n(n+1)/2] · [p(p+1)/2] = n(n+1)p(p+1)/4. In an n×n square grid: total rectangles = ⁿ⁺¹C₂² = [n(n+1)/2]² = n²(n+1)²/4. Total squares in n×n grid = n² + (n–1)² + … + 1² = n(n+1)(2n+1)/6. In n×p (n>p) grid: total squares = Σᵣ₌₁ᵖ (n+1–r)(p+1–r).

Parallelograms from intersecting lines (Permutations and Combinations): With n parallel lines in one set and m parallel lines in another set: number of parallelograms = ⁿC₂ × ᵐC₂ = [n(n–1)/2] · [m(m–1)/2]. Download the Free PDF for Permutations and Combinations for all geometric counting formulas with diagrams.

Exponent of Prime in n!, Trailing Zeros, and Rank of Word — Permutations and Combinations Applications

Why These Permutations and Combinations Applications Are Regularly Tested in JEE Main

Exponent of prime p in n! (Legendre's formula — Permutations and Combinations JEE Main): Eₚ(n!) = ⌊n/p⌋ + ⌊n/p²⌋ + ⌊n/p³⌋ + … (keep dividing n by powers of p and summing the floor values until p^k > n). This counts how many multiples of p, p², p³, … are in 1 to n, giving the total power of p in n!.

Trailing zeros of n! (Permutations and Combinations formula): A trailing zero comes from a factor of 10 = 2 × 5. Since there are always more factors of 2 than 5 in n!, the number of trailing zeros = E₅(n!) = ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + …. Example: trailing zeros in 100! = ⌊100/5⌋ + ⌊100/25⌋ = 20 + 4 = 24.

Special exponent results (Permutations and Combinations): E₆(n!) = E₃(n!) (since 6 = 2×3 and E₂ > E₃ always). E₁₀(n!) = E₅(n!) (same reason). Eₚ(ⁿPᵣ) = Eₚ(n!) – Eₚ((n–r)!). Eₚ(ⁿCᵣ) = Eₚ(n!) – Eₚ(r!) – Eₚ((n–r)!).

Sum of n-digit numbers (Permutations and Combinations): Sum of all numbers formed by n distinct non-zero digits (no repetition) = (sum of digits) × (n–1)! × (10ⁿ–1)/9. Example: sum of all 4-digit numbers using digits {1,2,3,4} (no repetition): sum = (1+2+3+4) × 3! × (10⁴–1)/9 = 10 × 6 × 1111 = 66660.

Rank of word in dictionary (Permutations and Combinations): Step 1: List all letters in alphabetical order. Step 2: For each position, count how many letters are smaller than the letter at that position (call it k). Step 3: Multiply k by (number of letters remaining)! — this gives how many words come before this letter choice. Step 4: Sum all such contributions and add 1 for the word itself = rank. For words with repeated letters: use factorial of remaining positions divided by factorials of repeated letters.

Solutions of linear equations (Permutations and Combinations — Stars and Bars): Number of non-negative integer solutions of x₁+x₂+…+xᵣ = n = ⁿ⁺ʳ⁻¹Cᵣ₋₁. Number of positive integer solutions (each xᵢ ≥ 1) = ⁿ⁻¹Cᵣ₋₁. With constraints (lᵢ ≤ xᵢ ≤ mᵢ): substitute yᵢ = xᵢ–lᵢ and use generating functions. Download the Free PDF for Permutations and Combinations for exponent formula examples and rank calculation worked problems.