Class XI · Chapter 6 · NCERT Mathematics

CHAPTER 06

Permutations & Combinations

The Art of Counting

Unlock the hidden order in arrangements — the mathematics of possibility itself.

\(nCr = n! / (r! · (n−r)!)\)
8 CBSE Marks
Difficulty
9 Topics
High JEE Weight

Topics Covered

9 key topics in this chapter

Fundamental Counting Principle
Factorial Notation
Permutations nPr
Permutations with Repetition
Circular Permutations
Combinations nCr
Combinations with Restrictions
Properties of nCr
Word Problems & Applications

Study Resources

𝑓 Key Formulae

Essential mathematical expressions for this chapter — understand derivations, not just results.

Factorial
\[n! = n \cdot (n-1) \cdot (n-2) \cdots 2 \cdot 1,\quad 0!=1\]
📌 Convention: 0! = 1
Permutation
\[^nP_r = \dfrac{n!}{(n-r)!}\]
📌 Ordered selection of r from n
Combination
\[^nC_r = \dfrac{n!}{r!\,(n-r)!}\]
📌 Unordered selection of r from n
Symmetry
\[^nC_r = {}^nC_{n-r}\]
📌 Choosing r is same as leaving n−r
Pascal's
\[^nC_r + {}^nC_{r-1} = {}^{n+1}C_r\]
📌 Row addition rule of Pascal's triangle
Circular Perm.
\[(n-1)!\text{ arrangements of }n\text{ objects in a circle}\]
📌 Fix one object, arrange the rest linearly
With Repetition
\[\text{Perm. with rep.} = \dfrac{n!}{p!\,q!\,r!\cdots}\]
📌 p,q,r… identical objects of each type

🎯 Exam-Ready Insights

Important points to remember — curated from CBSE Board question patterns.

01

CBSE 2-mark questions often ask: "In how many ways can 5 books be arranged on a shelf?" — apply nPr directly.

02

Distinguish P (order matters) vs C (order doesn't) from the problem context before calculating.

03

Circular arrangement of n distinct objects = (n−1)! — do NOT forget to fix one object.

04

Identical objects reduce the factorial: MATHEMATICS has 11 letters with repetitions — total = 11!/(2!·1!·…).

05

"At least one" problems: use complement — Total − None.

🏆 Competitive Exam Strategy

Targeted tips for JEE Main, JEE Advanced, NEET, BITSAT, and KVPY.

JEE Main

JEE Main loves "number of diagonals of a polygon" = ⁿC₂ − n, and "number of triangles formed by n points" = ⁿC₃ (when no 3 are collinear).

JEE Main

Gap method for non-adjacency: first arrange the un-restricted objects, then place the restricted ones in the gaps.

JEE Advanced

JEE Advanced tests multinomial coefficients and distributing identical/distinct objects into distinct/identical boxes — a rich topic.

BITSAT

BITSAT gives word-problems with conditions (vowels together, two specific people not together) — handle each condition separately.

⚠️ Common Mistakes to Avoid

Using ⁿPᵣ when the problem asks for selections (unordered) — always check if order matters.

Forgetting to account for identical elements in word-arrangement problems.

Double-counting in circular arrangements by not fixing one element.

Treating "at least one" as simple addition instead of using the complement rule.

💡 Key Takeaways

Permutation = arrangement (order matters); Combination = selection (order does not matter).

ⁿPᵣ = r! × ⁿCᵣ — every combination corresponds to r! permutations.

The Fundamental Counting Principle: if task A can be done in m ways and B in n ways independently, then together they can be done in m×n ways.

⁰Cᵣ = 0 for r>0; ⁿC₀ = ⁿCₙ = 1 always.

Repetition changes the formula significantly — always re-read the problem.

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