🎯 Key Points
- Permutations (order MATTERS): nPr = n!/(n-r)!; Combinations (order DOESN'T matter): nCr = n!/[r!(n-r)!] — the single biggest source of exam errors is picking the wrong one
- nCr = nC(n-r) (symmetry); nC0 = nCn = 1; Pascal's identity: nCr = (n-1)C(r-1) + (n-1)Cr
- Identical objects in a row: n!/(p!q!...) where p,q are counts of each repeated item
- nPr = r! × nCr — choose the group first (combination), then arrange it (factorial)
Permutations & Combinations
These are counting techniques used in probability and discrete mathematics.
Fundamental Counting Principle
Counting tree for selecting 2 items from {A, B, C} without repetition: 3 x 2 = 6 ordered arrangements.
If an event A can happen in m ways and B in n ways, then A AND B can happen in m × n ways (multiplication); A OR B in m + n ways (addition, if mutually exclusive).
Factorial
n! = n × (n-1) × ... × 2 × 1. By convention, 0! = 1.
Permutations (Arrangements: Order Matters)
- nPr = n! / (n-r)! = number of ways to arrange r objects from n distinct objects
- All n objects: nPn = n!
- With repetitions allowed: nʳ arrangements
- With identical objects: n! / (p! × q! × ...) where p, q are counts of identical objects
Combinations (Selections: Order Doesn't Matter)
- nCr = n! / [r!(n-r)!] = number of ways to choose r objects from n
- nCr = nCn-r (symmetry property)
- nC0 = nCn = 1
- Pascal's triangle: nCr = (n-1)Cr-1 + (n-1)Cr
Binomial Theorem
(a+b)ⁿ = Σ nCr × aⁿ⁻ʳ × bʳ, where sum runs r=0 to n.
General term: Tr+1 = nCr × aⁿ⁻ʳ × bʳ
Key Identities
- Sum of all combinations: 2ⁿ = nC0 + nC1 + ... + nCn
- nPr = r! × nCr (permutations = combinations × arrangements)
Permutations Under Conditions
- Objects together: tie the objects that must stay together into one block, arrange the blocks, then arrange within the block. For k objects together among n: (n − k + 1)! × k!.
- Objects never together (gap method): arrange the remaining objects first, then place the restricted objects into the gaps between them so they can never be adjacent.
- Fixed positions: if certain objects must occupy specific places (e.g. all vowels in even positions), fill those places first, then fill the remaining places.
Formation of Numbers
Digit problems are counting problems: decide how many choices each place value has, respecting the restrictions.
- The leading (left-most) digit usually cannot be 0, so it has fewer choices than the other places.
- "Digits not repeated" reduces the available pool by one for each place already filled; "digits may repeat" keeps the full pool for every place.
- Divisibility fixes the last digit(s): divisible by 2 → last digit even; by 5 → last digit 0 or 5; by 10 → last digit 0.
Division and Distribution into Groups
- The number of ways to divide (m + n) distinct objects into two unequal groups of sizes m and n is (m + n)!/(m! n!).
- To divide into two equal groups of size n each, use (2n)!/(n! n! 2!) — divide by 2! because the two equal groups can be swapped.
- If the groups are then assigned to distinct persons or positions, multiply by the factorial of the number of groups.
- Distributing n identical objects into r distinct boxes: each box ≥ 0 gives (n + r − 1)C(r − 1); each box ≥ 1 gives (n − 1)C(r − 1).
More Properties of Combinations
- nCr = nC(n − r): choosing r to keep is the same as choosing (n − r) to reject.
- nCr + nC(r − 1) = (n + 1)Cr (Pascal's rule, used to build Pascal's triangle).
- nCr / nC(r − 1) = (n − r + 1)/r, handy for the ratio of consecutive coefficients.
- If nCx = nCy then either x = y or x + y = n.
- nCr is greatest at the middle: r = n/2 (n even), or r = (n − 1)/2 and (n + 1)/2 (n odd).
Geometry Applications of Combinations
Selection problems on points and polygons are direct combination counts (assume no three points collinear unless stated).
- Straight lines through n points: nC2.
- Triangles from n points: nC3.
- Diagonals of a convex n-sided polygon: nC2 − n = n(n − 3)/2.
- If p of the n points are collinear, adjust for the lost lines/triangles: lines = nC2 − pC2 + 1, triangles = nC3 − pC3.
🚀 JEE Advanced Edge
Circular permutations: Arranging n distinct objects in a circle gives (n-1)! arrangements, not n!, because rotating the whole circle produces the same arrangement — fixing one object's position removes the redundant rotational symmetry. If clockwise and anticlockwise arrangements are considered identical (e.g. a necklace that can be flipped), divide further by 2: (n-1)!/2.
Distributing identical objects into groups (stars and bars): The number of ways to distribute n IDENTICAL items into r distinct groups (each group can be empty) is (n+r-1)C(r-1) — distinct from distributing DISTINCT items, which instead uses the multiplication principle (rⁿ ways if any group can hold any number of items).
Worked problem: In how many ways can 5 distinct books be arranged on a shelf such that 2 specific books always stay together (in either order)? Approach: Treat the 2 together-books as a single block, giving 4 units to arrange in 4! ways; the 2 books within the block can swap in 2! ways. Total = 4! × 2! = 24 × 2 = 48.
Worked Example: Letters Never Adjacent
In how many ways can the letters of PENCILS be arranged so that N and C are never adjacent?
Total arrangements of 7 distinct letters = 7! = 5040. Arrangements with N and C adjacent: treat NC as one block → 6 units arranged in 6! ways, with 2 internal arrangements for the block = 6! × 2 = 1440.
N and C not adjacent = 5040 − 1440 = 3600. The standard approach for "never together" is always: total − (treated as one block).
Worked Example: Committee with a Restriction
A committee of 5 is formed from 6 men and 4 women such that at least 2 women are included. How many ways?
Case 1 (exactly 2 women): C(4,2) × C(6,3) = 6 × 20 = 120. Case 2 (exactly 3 women): C(4,3) × C(6,2) = 4 × 15 = 60. Case 3 (exactly 4 women): C(4,4) × C(6,1) = 1 × 6 = 6.
Total = 120 + 60 + 6 = 186 ways. "At least k" problems: always split into cases (exactly k, k+1, …) rather than trying a single formula.