Permutations and Combinations

Permutations and Combinations

Board: CBSE | Class: Class 11

Comprehensive study notes for Permutations and Combinations by Ajay Yadav (Math King of Katargam). Master every concept with clear explanations, solved examples, and practice problems.

Key Concepts

Fundamental Principle of Counting

If one activity can be done in m ways and another in n ways, the two can be done together in m × n ways (multiplication principle). If either/or: m + n ways (addition principle).

Factorial

n! = n × (n-1) × (n-2) × … × 2 × 1. 0! = 1. n! = n × (n-1)!. Properties: n! = n(n-1)(n-2)! etc.

Permutations (nPr)

Number of ways to arrange r objects from n distinct objects: P(n,r) = n!/(n-r)!. Order matters. Example: 3 letter arrangements from 5 letters = 5P3 = 60.

Combinations (nCr)

Number of ways to select r objects from n distinct objects: C(n,r) = n!/[r!(n-r)!]. Order does NOT matter. nCr = nC(n-r). nC0 = nCn = 1.

Permutations with Repetition

If some objects are identical: Number of permutations of n objects where p are of one kind, q of another: n!/(p!q!…).

Circular Permutations

Arrangement around a circle: (n-1)! ways. If clockwise/anticlockwise are same: (n-1)!/2 ways.

Important Formulas

nPrP(n,r) = n!/(n-r)!
nCrC(n,r) = n!/[r!(n-r)!]
nCn = nC0nCn = nC0 = 1
nCr = nC(n-r)C(n,r) = C(n,n-r)
Circular(n-1)! (distinct arrangements)

Solved Examples

Example 1: How many 3-digit numbers can be formed from digits 1-6 without repetition?

Solution: P(6,3) = 6!/3! = 6×5×4 = 120.

Example 2: In how many ways can 4 books be arranged on a shelf?

Solution: 4! = 24 ways.

Example 3: How many committees of 3 can be formed from 10 people?

Solution: C(10,3) = 10!/(3!7!) = (10×9×8)/(3×2×1) = 120.

Practice Questions

  1. Find n if nP3 = 60.
  2. How many 4-letter words can be formed from “MATHEMATICS”?
  3. In how many ways can 6 people sit around a circular table?
  4. How many diagonals does a decagon have?
  5. Find n if nC10 = nC15.

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