Class 11 Maths NCERT Solutions Chapter 6 Permutations and Combinations | Boundless Maths
Chapter 6 Class 11 Maths NCERT Solutions · Unit II · Algebra

Class 11 Maths NCERT Solutions Chapter 6: Permutations and Combinations

Free, step-by-step NCERT Solutions for all four exercises of this chapter — the fundamental principle of counting, factorial notation, permutations of distinct and repeated objects (with and without repetition), and combinations — plus the Miscellaneous Exercise that mixes both ideas together. Solved the way CBSE awards marks, with the key formulas and the mistakes that cost students marks every year, right on this page.

Permutations and Combinations is the chapter where counting stops being "list everything out" and becomes a set of formulas — provided you can tell, for each question, whether order matters. It opens with the multiplication principle, a simple idea that answers surprisingly large counting problems without listing a single arrangement. From there it builds nPr for arranging things in order, and nCr for selecting things where order is irrelevant, along with the one identity, nPr = nCr × r!, that ties the two together.

5Exercises (incl. Misc.)
42Total Questions
2026-27CBSE Syllabus
100%Solved

Class 11 Maths NCERT Solutions Chapter 6 — Overview

This Class 11 Maths NCERT Solutions Chapter 6 hub covers Permutations and Combinations, the chapter that turns tedious listing into quick formulas. It opens with the fundamental principle of counting — if one event can happen in m ways and a following event in n ways, the two together happen in m × n ways — and uses it to build up factorial notation, since n! is just a compact way to write the product of counting down from n to 1.

From there, the chapter splits counting problems into two families. Permutations (nPr) count arrangements, where order matters — how many different 4-letter codes, how many ways to seat people, how many ways to form a number. Combinations (nCr) count selections, where order doesn't matter — how many possible committees, how many hands of cards. The Miscellaneous Exercise mixes both together in longer, exam-style problems, which is exactly the skill CBSE tests: correctly deciding which formula a question actually needs before reaching for it.

How the Chapter Builds

One Idea Leads to the Next

1

Fundamental Principle of Counting

The multiplication principle, m × n, that underlies every formula in this chapter. Exercise 6.1.

→
2

Factorial Notation

n! as a compact product, and why 0! = 1 by convention. Exercise 6.2.

→
3

Permutations

nPr, permutations with repetition, and arrangements when objects repeat. Exercise 6.3.

→
4

Combinations

nCr, and how it relates to nPr through r!. Exercise 6.4.

Quick Reference

Important Formulas — Chapter 6

Everything you need before you start solving. This is a summary for quick recall — the Formula Cards below has the full printable version for all of Permutations and Combinations.

Fundamental Principle of Counting (§6.2)

Multiplication principle

m \times n

If an event can occur in m ways, followed by another in n ways, the two together occur in m × n ways. Extends to any number of events.

Multiple stages

m \times n \times p

For three events in succession, multiply all three counts — the same reasoning as for two, applied stage by stage.

Either/or cases (addition)

\text{Total} = \text{Case 1} + \text{Case 2} + \ldots

When a signal or arrangement can have a variable number of parts (e.g. "at least 2 flags"), count each case separately and add the results.

Factorial Notation (§6.3.2)

Factorial

n! = 1\times 2\times 3\times \cdots \times n

Read as "n factorial." Also n! = n × (n − 1)!, which is what lets you cancel factorials in a ratio.

Zero factorial

0! = 1

Defined this way (not derived) so that the nPr and nCr formulas stay valid at r = 0 and r = n.

Permutations (§6.3)

Permutations, n distinct things taken r at a time

{}^{n}P_{r} = \dfrac{n!}{(n-r)!}, \ 0\le r\le n

Order matters, and no object repeats. This is the "fill r vacant places from n objects" formula.

Permutations with repetition allowed

n^{r}

Each of the r places can independently be filled in n ways — e.g. r-digit codes where digits may repeat.

Permutations of objects not all distinct

\dfrac{n!}{p_1!\,p_2!\,\cdots\,p_k!}

Used whenever a word or list has repeated letters/objects — divide out the repeats to avoid overcounting identical arrangements.

Combinations (§6.4)

Combinations, n distinct things taken r at a time

{}^{n}C_{r} = \dfrac{n!}{r!\,(n-r)!}, \ 0\le r\le n

Order doesn't matter — this counts selections, such as committees or hands of cards, not arrangements.

Link between P and C

{}^{n}P_{r} = {}^{n}C_{r}\times r!

Select r objects (nCr ways), then arrange them (r! ways) — together these give every permutation exactly once.

Complementary and Pascal's rule

{}^{n}C_{r} = {}^{n}C_{n-r} \qquad {}^{n}C_{r} + {}^{n}C_{r-1} = {}^{n+1}C_{r}

Choosing r objects is the same as rejecting (n − r) of them; the second identity is useful for simplifying sums of combinations.

Decision Guide

Which Formula Applies?

A quick way to decide, once you know what the question is actually asking for.

What the question is askingUse thisWhy
Two or more independent choices made one after anotherMultiplication principleMultiply the number of ways for each stage together (§6.2).
Arranging r distinct objects out of n, no repetition, order mattersPermutation, nPrEvery different order is a different outcome (§6.3.1).
Forming codes/numbers where digits or letters can repeatnrEach position is filled independently, with no restriction from earlier choices (§6.3.1, Theorem 2).
Arranging all letters of a word with repeated lettersn!/(p₁!p₂!…)Divide by the factorial of each repeat count to avoid counting identical arrangements twice (§6.3.4).
Selecting a committee, team or hand — order irrelevantCombination, nCrThe same group in a different order is still the same selection (§6.4).
Selecting a group and then also assigning roles/order within itnCr × r!First select, then arrange — this recovers nPr (§6.4, Theorem 5).
Letters/objects must stay together in a blockGroup as one unit, then multiply by internal arrangementsTreat the block as a single object, arrange it with the rest, then arrange inside the block separately (§6.3.4).
"At least" or "at most" conditions on a selectionSum the valid cases, or subtract the complement from the totalBreak into disjoint cases (exactly 0, exactly 1, …) and add, or find what's excluded and subtract it from the unrestricted count.
Avoid These

Common Mistakes to Avoid in This Chapter

Drawn from where students actually lose marks across all four exercises.

  • Using nPr when the question is actually a selection — always ask first whether rearranging the same objects should count as a new outcome. If not, it's a combination, not a permutation.
  • Forgetting to divide by p! when objects repeat — arranging the letters of a word like INSTITUTE or MISSISSIPPI without dividing by the factorial of each repeated letter's count overcounts identical-looking arrangements many times over.
  • Letting a leading digit be zero — in "how many 4-digit numbers" type questions, the first (leftmost) place cannot be filled with 0, so it must be counted separately from the remaining places.
  • Forgetting the internal arrangement when objects must stay together — treating a block of letters as one unit gives the arrangements of the block with everything else, but the letters inside the block can usually still be reordered among themselves, and that factor is often missed.
  • Mishandling "at least" and "at most" conditions — these need either a sum over multiple valid cases or a subtraction from the total; picking only one case, or forgetting a boundary case (like all objects being of one kind), is a frequent error.
  • Applying 0! = 1 inconsistently, or forgetting it entirely — this shows up constantly in ratio-based questions ("find n if…", "find r if…") and in edge cases where r = 0 or r = n.
  • Skipping the r! after selecting with nCr — whenever a selection is followed by assigning positions, roles, or an order, multiply the selection count by the number of arrangements of what was selected.
Solve Chapter-Wise

Class 11 Maths NCERT Solutions Chapter 6 — Choose an Exercise

6.1

Exercise 6.1

The fundamental principle of counting, applied to digits, letters, coins and flags · 6 questions

Solve Exercise 6.1 →
6.2

Exercise 6.2

Factorial notation — evaluating, simplifying, and solving for unknowns in factorial ratios · 5 questions

Solve Exercise 6.2 →
6.3

Exercise 6.3

nPr, word arrangements, and permutations of objects that are not all distinct · 11 questions

Solve Exercise 6.3 →
6.4

Exercise 6.4

nCr — committees, chords, card hands, and selections with fixed conditions · 9 questions

Solve Exercise 6.4 →
M

Miscellaneous Exercise

Longer problems mixing permutations and combinations together, exactly as CBSE tests them · 11 questions

Solve Miscellaneous →

📐 Keep the Formulas Handy

Every formula for Permutations and Combinations — plus every other Class 11 Maths chapter — in one printable PDF.

Get Formula Cards →
Common Questions

Frequently Asked Questions

Quick answers from Class 11 Maths NCERT Solutions Chapter 6, Permutations and Combinations.

How many exercises are there in Chapter 6, Permutations and Combinations?
There are four main exercises — 6.1 (Fundamental Principle of Counting, 6 questions), 6.2 (Factorial Notation, 5 questions), 6.3 (Permutations, 11 questions) and 6.4 (Combinations, 9 questions) — plus a Miscellaneous Exercise of 11 questions combining both ideas, totalling 42 questions.
What is the difference between a permutation and a combination?
A permutation is an arrangement in a definite order, so changing the order of the same objects gives a different permutation. A combination is a selection where order is irrelevant, so the same set of objects in any order counts only once. Arranging 3 letters out of 5 is a permutation problem; picking a 3-person committee out of 5 people is a combination problem.
Why is 0! defined as 1?
0! = 1 is a convention, not something proved from the definition n! = 1 × 2 × 3 × ... × n. It is fixed this way so that formulas like nPr = n!/(n−r)! and nCr = n!/(r!(n−r)!) continue to hold at the boundary case r = n, and so that nC0 = 1 correctly reflects that there is exactly one way to select nothing at all from a set — leaving every object behind.
When do I multiply nCr by r! in a problem?
Multiply by r! whenever a problem asks you to both select a group of r objects and then arrange that group in a definite order. First use nCr to count the selection, then multiply by r! for the arrangements of whatever was selected. Skipping the r! is a common cause of answers coming out exactly a factor of r! too small.
Where can I find the official NCERT textbook for this chapter?
Permutations and Combinations is Chapter 6 of the NCERT Class 11 Mathematics textbook, published by the National Council of Educational Research and Training (NCERT) and prescribed by CBSE. You can download the official textbook PDF directly from ncert.nic.in, NCERT's official website — the solutions on this page follow the exercises exactly as they appear there.
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