JEE MATH APEX
JEE MATH APEX
JEE MATH APEX
JEE MATH APEX
JEE MATH APEX
JEE MATH APEX
JEE MATH APEX
JEE MATH APEX
JEE MATH APEX
JEE MATH APEX
PERMUTATIONS AND COMBINATIONS
— JEE MATH APEX —
JEE MATH APEX
1. Fundamental Counting Principle
📌 Multiplication & Addition Principle
• Multiplication Principle: If task 1 can be done in m ways and task 2 in n ways, then both can be done in m × n ways
• Addition Principle: If task 1 can be done in m ways and task 2 in n ways (mutually exclusive), then either can be done in m + n ways
• If m tasks with n₁, n₂, …, nₘ ways: Total = n₁ × n₂ × … × nₘ
• AND operation → Multiplication
• OR operation → Addition
• Addition Principle: If task 1 can be done in m ways and task 2 in n ways (mutually exclusive), then either can be done in m + n ways
• If m tasks with n₁, n₂, …, nₘ ways: Total = n₁ × n₂ × … × nₘ
• AND operation → Multiplication
• OR operation → Addition
JEE MATH APEX
2. Permutation Formula
📌 nPr Formula
• Permutation of n different objects taken r at a time: P(n,r) = nPr = n!/(n−r)!
• nPr = n(n−1)(n−2)…(n−r+1)
• nPn = n!
• nP0 = 1
• nP1 = n
• 0! = 1, 1! = 1
• nPr = n(n−1)(n−2)…(n−r+1)
• nPn = n!
• nP0 = 1
• nP1 = n
• 0! = 1, 1! = 1
JEE MATH APEX
3. Combination Formula
📌 nCr Formula
• Combination of n different objects taken r at a time: C(n,r) = nCr = n!/(r!(n−r)!)
• nCr = nPr / r!
• nC0 = 1, nCn = 1
• nC1 = n
• nCr = nC(n−r)
• nCr = nPr / r!
• nC0 = 1, nCn = 1
• nC1 = n
• nCr = nC(n−r)
JEE MATH APEX
4. Properties of nCr
📌 Binomial Coefficient Properties
• nCr = nC(n−r) (Symmetry property)
• nCr + nC(r−1) = (n+1)Cr (Pascal’s rule)
• nCr / nC(r−1) = (n−r+1)/r
• nCr = (n/r) × (n−1)C(r−1)
• nC0 + nC1 + nC2 + … + nCn = 2n
• nC0 + nC2 + nC4 + … = nC1 + nC3 + nC5 + … = 2n−1
• nCr + nC(r−1) = (n+1)Cr (Pascal’s rule)
• nCr / nC(r−1) = (n−r+1)/r
• nCr = (n/r) × (n−1)C(r−1)
• nC0 + nC1 + nC2 + … + nCn = 2n
• nC0 + nC2 + nC4 + … = nC1 + nC3 + nC5 + … = 2n−1
JEE MATH APEX
5. Permutations with Repetition
📌 Repeated Objects
• n objects with p identical of one type, q identical of another type:
Arrangements = n! / (p! × q!)
• General: n objects with n₁, n₂, …, nₖ identical objects:
Arrangements = n! / (n₁! × n₂! × … × nₖ!)
• Permutations when repetition is allowed: nr
• Arrangement of n different objects taken all at a time: n!
Arrangements = n! / (p! × q!)
• General: n objects with n₁, n₂, …, nₖ identical objects:
Arrangements = n! / (n₁! × n₂! × … × nₖ!)
• Permutations when repetition is allowed: nr
• Arrangement of n different objects taken all at a time: n!
JEE MATH APEX
6. Circular Permutations
📌 Circular Arrangements
• Arrangement of n distinct objects in a circle: (n−1)!
• If clockwise and anticlockwise arrangements are same (necklace): (n−1)!/2
• Arrangement of n distinct objects in a circle with a fixed point: (n−1)!
• Circular permutation of n objects taken r at a time: nPr / r
• If clockwise and anticlockwise arrangements are same (necklace): (n−1)!/2
• Arrangement of n distinct objects in a circle with a fixed point: (n−1)!
• Circular permutation of n objects taken r at a time: nPr / r
JEE MATH APEX
7. Selection with Restrictions
📌 Conditional Selection
• Selecting r objects from n with k specific objects always included: (n−k)C(r−k)
• Selecting r objects from n with k specific objects always excluded: (n−k)Cr
• Selecting at least 1 object from n distinct objects: 2n − 1
• Selecting at least 1 object from n identical objects: n ways
• Total combinations from p distinct + q identical: (2p) × (q + 1) − 1
• Selecting r objects from n with k specific objects always excluded: (n−k)Cr
• Selecting at least 1 object from n distinct objects: 2n − 1
• Selecting at least 1 object from n identical objects: n ways
• Total combinations from p distinct + q identical: (2p) × (q + 1) − 1
JEE MATH APEX
8. Distribution of Objects
📌 Grouping & Distribution
• Distributing n distinct objects among r persons (each can get any number): rn
• Distributing n identical objects among r persons: (n+r−1)C(r−1)
• Dividing n distinct objects into groups of sizes n₁, n₂, …, nₖ:
n! / (n₁! × n₂! × … × nₖ!) (if groups are distinct)
• If groups are identical and of equal size, divide by factorial of number of groups
• Distributing n identical objects among r persons: (n+r−1)C(r−1)
• Dividing n distinct objects into groups of sizes n₁, n₂, …, nₖ:
n! / (n₁! × n₂! × … × nₖ!) (if groups are distinct)
• If groups are identical and of equal size, divide by factorial of number of groups
JEE MATH APEX
9. Arrangement with Restrictions
📌 Together / Not Together / Specific Position
• k objects always together: Treat as one unit → (n−k+1)! × k!
• k objects never together: Total − Together = n! − (n−k+1)! × k!
• n objects with specific object at fixed position: (n−1)!
• n objects with two specific objects adjacent: 2 × (n−1)!
• Arrangement with alternate positions: Use gap method
• k objects never together: Total − Together = n! − (n−k+1)! × k!
• n objects with specific object at fixed position: (n−1)!
• n objects with two specific objects adjacent: 2 × (n−1)!
• Arrangement with alternate positions: Use gap method
JEE MATH APEX
10. Rank of a Word
📌 Dictionary Order Problems
• To find rank of a word in dictionary:
Step 1: Arrange letters alphabetically
Step 2: Count words before given word
Step 3: Rank = (words before) + 1
• Formula: Rank = 1 + Σ [Number of letters less than current × (remaining letters)!/(repetitions!)]
Step 1: Arrange letters alphabetically
Step 2: Count words before given word
Step 3: Rank = (words before) + 1
• Formula: Rank = 1 + Σ [Number of letters less than current × (remaining letters)!/(repetitions!)]
JEE MATH APEX
11. Gap Method
📌 Arrangement with No Two Adjacent
• Arrange m objects so that no two of k specific objects are together:
Step 1: Arrange (m−k) objects → (m−k)!
Step 2: Create gaps → (m−k+1) gaps
Step 3: Select k gaps → (m−k+1)Ck × k!
• Total: (m−k)! × (m−k+1)Ck × k!
• Used for: Alternating arrangement, no two together problems
Step 1: Arrange (m−k) objects → (m−k)!
Step 2: Create gaps → (m−k+1) gaps
Step 3: Select k gaps → (m−k+1)Ck × k!
• Total: (m−k)! × (m−k+1)Ck × k!
• Used for: Alternating arrangement, no two together problems
JEE MATH APEX
12. Number of Divisors
📌 Divisors of a Number
• If N = p₁a₁ × p₂a₂ × … × pₖaₖ (prime factorisation)
• Total number of divisors: (a₁+1)(a₂+1)…(aₖ+1)
• Sum of divisors: [(p₁a₁+1−1)/(p₁−1)] × … × [(pₖaₖ+1−1)/(pₖ−1)]
• Product of divisors: Nτ(N)/2 (where τ is number of divisors)
• Number of even divisors: Count with at least one factor of 2
• Total number of divisors: (a₁+1)(a₂+1)…(aₖ+1)
• Sum of divisors: [(p₁a₁+1−1)/(p₁−1)] × … × [(pₖaₖ+1−1)/(pₖ−1)]
• Product of divisors: Nτ(N)/2 (where τ is number of divisors)
• Number of even divisors: Count with at least one factor of 2
JEE MATH APEX
13. Sum of Numbers Formed
📌 Sum of All Numbers Formed by Digits
• Sum of all n-digit numbers formed by n distinct digits (no repetition):
Sum = (n−1)! × (Sum of digits) × (111…n times)
• Sum of all r-digit numbers from n distinct digits:
Sum = nPr/n × (Sum of digits) × (111…r times)
• If repetition allowed: Sum = nr−1 × (Sum of digits) × (111…r times)
Sum = (n−1)! × (Sum of digits) × (111…n times)
• Sum of all r-digit numbers from n distinct digits:
Sum = nPr/n × (Sum of digits) × (111…r times)
• If repetition allowed: Sum = nr−1 × (Sum of digits) × (111…r times)
JEE MATH APEX
14. Derangement
📌 No Object at Its Original Position
• Derangement formula: Dn = n! [1 − 1/1! + 1/2! − 1/3! + … + (−1)ⁿ/n!]
• D₁ = 0
• D₂ = 1
• D₃ = 2
• D₄ = 9
• Recurrence: Dn = (n−1)[Dn−1 + Dn−2]
• D₁ = 0
• D₂ = 1
• D₃ = 2
• D₄ = 9
• Recurrence: Dn = (n−1)[Dn−1 + Dn−2]
JEE MATH APEX
15. Multinomial Theorem
📌 Number of Terms & Coefficient
• Number of terms in expansion of (x₁ + x₂ + … + xₖ)ⁿ: (n+k−1)C(k−1)
• Coefficient of x₁a₁x₂a₂…xₖaₖ: n!/(a₁! × a₂! × … × aₖ!)
• Where a₁ + a₂ + … + aₖ = n
• Coefficient of x₁a₁x₂a₂…xₖaₖ: n!/(a₁! × a₂! × … × aₖ!)
• Where a₁ + a₂ + … + aₖ = n
✅ PERMUTATIONS AND COMBINATIONS ✅
— JEE MATH APEX —
— JEE MATH APEX —
JEE MATH APEX 
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