Chapter 6 of 14

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Odisha (BSE)Class 11 Mathematics★ Unit II · part of the 25-mark Algebra block shared with Complex Numbers and Quadratic Equations, Linear Inequalities, Binomial Theorem, and Sequences and Series — CBSE states no chapter-wise split

Permutations and Combinations

How many 3-digit sequences would you need to try on a forgotten suitcase lock? This chapter builds the counting tools to answer that without listing every possibility — the multiplication principle, factorials, permutations for when order matters, and combinations for when it doesn't. Unlike the last few chapters, almost everything here is still summatively examined — the one exception is circular arrangements, a coaching-material staple that isn't in the current book at all. CBSE Class 11 Mathematics Chapter 6.

⏱ 9 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Apply the fundamental principle of counting to multi-stage problems, and know when to multiply versus add across cases
2Compute factorials and use nPr for arrangements of distinct objects, with or without repetition allowed
3Count arrangements of objects that are not all distinct, and handle 'keep together' and 'keep apart' restrictions
4Compute nCr for selections and apply nCr = nC(n-r) and Pascal's identity
5Solve 'at least' and 'at most' selection problems by splitting into cases and adding

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Why this chapter matters

This chapter builds the counting tools used everywhere probability is computed: the multiplication principle for sequential choices, permutations for arrangements where order matters, and combinations for selections where it doesn't. Unlike several neighbouring chapters, almost none of its content has been pushed to formative-only status, and it feeds directly into the Binomial Theorem via Pascal's identity.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Basic set notation, since combinations count subsets of a fixed size

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Comfort with algebraic manipulation of factorials and ratios, used throughout the nPr and nCr formulas

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The habit of systematic case-listing, extended here from simple counting into formal combinatorics

Permutations and Combinations

1. Check this before you revise anything

Good news first: unlike Complex Numbers, Linear Inequalities, or Trigonometric Functions, almost nothing in this chapter has been pushed to formative-only status. CBSE's syllabus line reads in full: "Fundamental principle of counting. Factorial n. Permutations and combinations, derivation of formulae for $^{n} P_{r}$ and $^{n} C_{r}$ and their connections, simple applications" — and every part of that is summatively examined, matching the current book closely.

One coaching-material staple genuinely isn't here, though. Circular permutations (arranging people around a table, where rotations count as the same arrangement) do not appear anywhere in the 2026-27 chapter — not in the theorems, not in any of the 42 questions across its five exercises. If a worksheet asks about seating people in a circle, it is drilling content from outside this chapter entirely, not a formative-only carve-out like recent chapters — it simply was never part of this book.

2. What this chapter covers

Textbook section	Topic
6.2	The fundamental principle of counting (multiplication principle)
6.3	Permutations — factorial notation, $^{n} P_{r}$ , arrangements with repetition or identical objects
6.4	Combinations — $^{n} C_{r}$ , its connection to $^{n} P_{r}$ , and Pascal-style identities

3. The fundamental principle of counting

If one event can happen in $m$ ways, and a second (independent) event can happen in $n$ ways, the two together can happen in $m \times n$ ways. This extends to any number of events: $m_{1} \times m_{2} \times m_{3} \times \dots$ .

Worked, mirroring the textbook's own Example 2. Given 4 flags of different colours, how many 2-flag signals (one above the other) can be made? The upper position has 4 choices; for each, the lower position has 3 remaining choices: $4 \times 3 = 12$ .

Worked, mirroring the textbook's own Example 4. How many signals can be made using at least 2 of 5 available flags? A signal could use 2, 3, 4, or 5 flags — count each case separately, then add them, since a signal can't simultaneously have 2 flags and 3 flags: $5 \times 4 + 5 \times 4 \times 3 + 5 \times 4 \times 3 \times 2 + 5 \times 4 \times 3 \times 2 \times 1 = 20 + 60 + 120 + 120 = 320$ .

This is the chapter's first real distinction, and it matters throughout: multiply for "this AND then that" (sequential, independent choices); add for "this OR that" (separate, mutually exclusive cases).

4. Factorial notation, and permutations of distinct objects

$n! = n \times (n - 1) \times (n - 2) \times \dots \times 2 \times 1$ , with $0! = 1$ by definition — a convention chosen specifically so the formulas below work at their boundary cases.

A permutation is an arrangement in a definite order. The number of permutations of $n$ distinct objects taken $r$ at a time is:

$^{n} P_{r} = \frac{n !}{( n - r )!}, 0 \leq r \leq n$

This comes directly from the multiplication principle: the first of $r$ positions has $n$ choices, the second has $n - 1$ remaining, ..., the $r$ -th has $n - (r - 1)$ remaining. When $r = n$ : $^{n} P_{n} = n!$ . When $r = 0$ : $^{n} P_{0} = 1$ (there's exactly one way to arrange nothing — leave everything behind).

Worked, mirroring the textbook's own Example 11. How many numbers between 100 and 1000 use the digits $0, 1, 2, 3, 4, 5$ without repetition? Every such number has exactly 3 digits, so start with $^{6} P_{3} = 120$ arrangements of 3 digits from 6 — but this overcounts, including arrangements like $042$ that aren't really 3-digit numbers (leading zero). Fixing $0$ in the hundreds place and arranging 2 digits from the remaining 5 gives $^{5} P_{2} = 20$ such invalid cases. The answer is $120 - 20 = 100$ .

When repetition is allowed, the count becomes $n^{r}$ instead — each of the $r$ positions independently has all $n$ choices again.

5. Permutations when objects repeat

If some of the $n$ objects are identical, straightforward arrangement over-counts: swapping two identical objects doesn't create a new arrangement, but the $n!$ formula would count it as if it did. For $p_{1}$ objects of one kind, $p_{2}$ of a second kind, and so on:

$Arrangements = \frac{n !}{p _{1} ! p _{2} ! \dots p _{k} !}$

Worked, mirroring the textbook's own Example 9. ALLAHABAD has 9 letters: 4 A's, 2 L's, the rest distinct. Arrangements: $\frac{9 !}{4 ! 2 !} = 7560$ .

Worked, mirroring the textbook's own Example 16. INDEPENDENCE has 12 letters (N×3, E×4, D×2, rest distinct). Total arrangements: $\frac{12 !}{3 ! 4 ! 2 !} = 1663200$ . Restricted counts build on this base rather than starting over. Words starting with P: fix P, arrange the remaining 11 letters (still N×3, E×4, D×2): $\frac{11 !}{3 ! 4 ! 2 !} = 138600$ .

Words where all 5 vowels (E,E,E,E,I) stay together: treat them as one block, giving 8 objects (block + 7 consonants, still N×3, D×2) to arrange, times the vowels' own internal arrangements: $\frac{8 !}{3 ! 2 !} \times \frac{5 !}{4 !} = 16800$ . Vowels never all together is then found by subtracting from the unrestricted total: $1663200 - 16800 = 1646400$ .

6. Permutations with position restrictions

Worked, mirroring the textbook's own Example 14. DAUGHTER has 8 distinct letters, 3 of them vowels (A, U, E). Arrangements with all vowels together: treat AUE as one block, giving 6 objects to arrange ( $6!$ ), times the vowels' internal order ( $3!$ ): $6! \times 3! = 4320$ . Arrangements where the vowels are never all together: total unrestricted arrangements minus the "together" count: $8! - 4320 = 36000$ .

Worked, mirroring the textbook's own Example 24 — a genuinely different technique, for keeping objects apart rather than together. Seat 5 girls and 3 boys in a row so no two boys sit together. Seat the girls first: $5!$ ways. This creates 6 gaps (including both ends: $_G_G_G_G_G_$ ) — placing the 3 boys into 3 of these 6 gaps, order mattering, is $^{6} P_{3}$ . Total: $5! \times^{6} P_{3} = 120 \times 120 = 14400$ .

The "gap method" is the standard tool whenever a problem says objects must NOT be adjacent — arrange the unrestricted group first, then place the restricted objects into the gaps this creates.

7. Combinations

A combination is a selection where order doesn't matter — choosing a 3-person committee, not assigning who's president. Since every combination of $r$ objects can be internally reordered in $r!$ ways to produce a permutation:

$^{n} P_{r} =^{n} C_{r} \times r! ⟹^{n} C_{r} = \frac{n !}{r ! ( n - r )!}, 0 \leq r \leq n$

Two identities follow directly: $^{n} C_{r} =^{n} C_{n - r}$ (choosing $r$ to include is the same count as choosing $n - r$ to exclude), and Pascal's identity, $^{n} C_{r - 1} +^{n} C_{r} =^{(n + 1)} C_{r}$ — the rule behind every entry of Pascal's triangle, and the one this chapter hands directly to the Binomial Theorem next.

Worked, mirroring the textbook's own Example 18. A 3-person committee from 2 men and 3 women: $^{5} C_{3} = 10$ total ways. Restricted to exactly 1 man and 2 women: $^{2} C_{1} \times^{3} C_{2} = 2 \times 3 = 6$ .

Worked, mirroring the textbook's own Example 19(v). Choosing 4 cards of the same colour from a standard deck (26 red, 26 black): choose all 4 from red, or all 4 from black — cases are added, not multiplied, since a hand can't be both: $^{26} C_{4} +^{26} C_{4} = 29900$ .

8. "At least" and "at most" — splitting into cases

Worked, mirroring the textbook's own Example 21. From 4 girls and 7 boys, form a 5-member team with at least 3 girls. Since the group has only 4 girls, the only possible splits are 3 girls+2 boys, or 4 girls+1 boy — never "5 girls," which doesn't exist here. $^{4} C_{3} \times^{7} C_{2} +^{4} C_{4} \times^{7} C_{1} = 84 + 7 = 91$ .

This case-by-case approach — list every valid split, compute each with combinations, then add — is the standard method for every "at least" or "at most" selection problem, exactly parallel to how Section 3 handled "at least 2 flags" by adding across cases.

Summary

Multiply for sequential independent choices ("this AND then that"); add for separate, mutually exclusive cases ("this OR that").
$^{n} P_{r} = \frac{n !}{( n - r )!}$ counts arrangements of $r$ objects from $n$ distinct ones, order mattering; with repetition allowed, it becomes $n^{r}$ .
Objects that repeat divide out their own internal rearrangements: $n! / (p_{1}! p_{2}! \dots)$ .
"Keep together" problems glue the group into one block; "keep apart" problems arrange everything else first and place the restricted objects into the gaps.
$^{n} C_{r} = \frac{n !}{r ! ( n - r )!} = \frac{^{n} P _{r}}{r !}$ counts selections, order not mattering; $^{n} C_{r} =^{n} C_{n - r}$ , and Pascal's identity $^{n} C_{r - 1} +^{n} C_{r} =^{(n + 1)} C_{r}$ carries directly into the Binomial Theorem.
"At least"/"at most" problems are solved by listing every valid case, computing each with combinations, and adding — never by a single shortcut formula.
Circular permutations are not part of this chapter's current content at all, despite being a common coaching-material addition for this topic.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Fundamental principle of counting

If one event happens in m ways and a second happens in n ways, both together happen in m x n ways

Multiply for sequential/independent choices; add instead for separate, mutually exclusive cases

Factorial

n! = n(n-1)(n-2)...2.1; 0! = 1

0! = 1 is a convention chosen so nPr and nCr work correctly at r=0 and r=n

Permutations of distinct objects

nPr = n!/(n-r)!, for 0 <= r <= n

Counts ordered arrangements of r objects chosen from n distinct ones

Permutations with repetition allowed

n^r

Each of the r positions independently has all n choices, since objects can repeat

Permutations with identical objects

n! / (p1! p2! ... pk!)

Divide by the factorial of each group of identical objects to remove indistinguishable repeats

Combinations

nCr = n!/[r!(n-r)!] = nPr/r!

Counts unordered selections; every combination corresponds to r! permutations

Symmetry of combinations

nCr = nC(n-r)

Choosing r objects to include is the same count as choosing n-r objects to exclude

Pascal's identity

nC(r-1) + nCr = (n+1)Cr

The rule behind every entry of Pascal's triangle; carries directly into the Binomial Theorem

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Using nPr for a selection problem, or nCr for an arrangement problem

✓ Ask first: does swapping the order of the chosen items produce a different outcome? If yes (seating order, who holds which office) use nPr; if no (a team, a committee) use nCr.

WATCH OUT

✗ Treating 0! as 0 instead of 1

✓ 0! = 1 by definition — there is exactly one way to arrange or select nothing at all. Using 0 breaks nP0 and nC0, both of which must equal 1.

WATCH OUT

✗ Multiplying across cases that are actually mutually exclusive (or adding across steps that actually happen together)

✓ Multiply for 'this AND then that' (sequential/independent steps). Add for 'this OR that' (separate cases that can't both occur, like choosing all-red or all-black cards).

WATCH OUT

✗ Forgetting to divide by the factorial of each repeated letter/object when counting arrangements

✓ For a word like INDEPENDENCE, dividing by 12! alone overcounts massively — divide by 3! (three Ns), 4! (four Es), and 2! (two Ds) as well: 12!/(3!4!2!).

WATCH OUT

✗ Expecting circular-arrangement questions (seating around a table) as part of this chapter

✓ Circular permutations are not part of the current 2026-27 NCERT chapter at all — none of its 42 questions ask for one. It is coaching-material content from outside this specific book.

WATCH OUT

✗ Computing a 'never together' or 'not adjacent' count directly instead of via subtraction or the gap method

✓ For 'never together', compute (total unrestricted arrangements) minus (arrangements with them together) — direct case-listing is far more error-prone. For 'not adjacent', arrange everything else first, then place the restricted objects into the gaps this creates.

WATCH OUT

✗ In an 'at least' problem, forgetting the case where the minimum required amount is drawn entirely from one group (e.g. all 5 team members being girls, if the group is large enough)

✓ List every valid split systematically by the count of one category (0, 1, 2, ... up to whatever the group sizes allow), not just the 'obvious' middle cases.

WATCH OUT

✗ Assuming nCr always increases as r increases

✓ nCr increases up to r=n/2 and then decreases back down to nCn=1 — it is symmetric, not monotonic, following directly from nCr=nC(n-r).

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex 6.1

Exercise 6.1

Fundamental principle of counting — multi-stage counting problems using the multiplication principle

Questions

Ex 6.2

Exercise 6.2

Factorial notation — evaluating factorial expressions and solving simple factorial equations

Questions

Ex 6.3

Exercise 6.3

Permutations — nPr for distinct objects, solving for n or r, and arrangements with 'keep together'/'never together' restrictions

Questions

Ex 6.4

Exercise 6.4

Combinations — nCr for selections, chords/handshake-style counting, and card and committee selection problems

Questions

Miscellaneous Ex

Miscellaneous Exercise

Mixed permutation and combination problems, including dictionary-order word ranking, committee selection with case splits, and arrangements with positional constraints

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Permutations and Combinations?

10 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

10 questions~7 min worth ~25 marks in Odisha (BSE) exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Multiply for sequential/independent choices; add for separate, mutually exclusive cases
•n! = n(n-1)...1, with 0!=1 by convention
•nPr = n!/(n-r)! counts ordered arrangements; with repetition allowed, use n^r instead
•Identical objects divide out their own repeats: n!/(p1!p2!...)
•'Keep together' problems glue the group into one block; 'not adjacent' problems use the gap method — arrange the rest first, then place the restricted objects into the gaps
•nCr = n!/[r!(n-r)!] = nPr/r! counts unordered selections
•nCr = nC(n-r); Pascal's identity nC(r-1)+nCr = (n+1)Cr feeds directly into the Binomial Theorem
•'At least'/'at most' problems are solved by listing every valid case and adding, never by a single shortcut formula
•Circular permutations are not part of this chapter's current NCERT content

Odisha (BSE) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: Part of Unit II's 25-mark Algebra block (no chapter-wise split, per CBSE)

Question type	Marks each	Typical count	What it tests
Fundamental counting principle and factorial notation	2-3	1	Multi-stage counting via the multiplication principle, factorial evaluation and simple factorial equations
Permutations distinct repeated and restricted arrangements	3-5	1-2	nPr for distinct objects, arrangements with identical objects, 'keep together'/'never together'/'not adjacent' restrictions
Combinations and selection problems	3-4	1	nCr direct application, chord/handshake-style counting, Pascal's identity and nCr=nC(n-r)
At least at most case splitting problems	4-6	0-1	Committee or card-hand selection with a minimum or maximum count on one category, solved by listing and adding cases

Prep strategy

Before writing any formula, decide explicitly whether order matters — this single question resolves nearly every P-vs-C confusion
Memorise small factorials (up to 8! or so) to speed up direct computation without a calculator
For 'at least'/'at most' problems, write out every valid case in a small table before computing anything — skipping this step is the most common source of missed cases

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Lottery and lock combinations

A 4-digit suitcase lock with no repeated digits has 10P4 possible codes — exactly the multiplication-principle reasoning this chapter's own opening example uses.

Tournament and match scheduling

Round-robin scheduling among n teams, where each pair plays exactly once, needs nC2 matches — a direct combination, since 'Team A vs Team B' is the same match as 'Team B vs Team A'.

Card games and probability

The number of possible 5-card poker hands from a 52-card deck is 52C5 — the starting point for every probability calculation in the very next chapter.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

State explicitly whether a problem needs a permutation or combination before writing any formula — this alone earns method marks even if arithmetic errors follow

For 'at least'/'at most' problems, write out every valid case in a small table first, then compute and sum

For restricted-arrangement problems, decide up front: is this a 'keep together' (block method) or 'keep apart' (gap method) situation?

Double-check whether repetition is allowed before choosing between nPr and n^r — the question usually states this explicitly, and it's easy to miss

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

Combinatorial proof of the Vandermonde identity: sum over k of (mCk)(nC(r-k)) = (m+n)Cr, provable by a direct counting argument (splitting an (m+n)-element set into two parts) rather than algebra

STRETCH

Derangements: the number of permutations of n objects with no object in its original position, D(n) = n!(1 - 1/1! + 1/2! - ... + (-1)^n/n!) — a classic extension using inclusion-exclusion

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JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainDictionary-order ranking of a word's permutationsSystematic case elimination

If all the letters of the word DELHI are arranged in dictionary order, how many words come before DELHI itself?

Stuck? Show the approach

Fix letters one position at a time in the order they appear in DELHI, and at each position count how many of the remaining letters are alphabetically smaller than the one actually used — each such smaller letter starts a whole block of words that come earlier.

Show the full solution

Alphabetical order of D,E,H,I,L. Position 1 (D): no letter smaller than D among {D,E,H,I,L}, so 0 words. Position 2, with D fixed (E): among remaining {E,H,I,L}, none smaller than E, so 0 words. Position 3, with D,E fixed (L): among remaining {H,I,L}, two letters (H,I) are smaller than L, each contributing 2! = 2 words: 2x2=4. Position 4, with D,E,L fixed (H): among remaining {H,I}, none smaller than H, so 0. Position 5 is forced (I), giving DELHI itself, not a word before it.

Answer: 4

The trap

Forgetting to multiply each 'smaller letter found' by the factorial of how many positions remain after it is the most common slip — each smaller choice at an early position unlocks a whole block of later arrangements, not just one word.

JEE MainSeating with a no-two-adjacent restrictionGap method

In how many ways can 6 boys and 4 girls be arranged in a row so that no two girls sit together?

Stuck? Show the approach

Seat the unrestricted, larger group first, count the gaps this creates (including both ends), then place the restricted group into those gaps using a permutation, since which girl sits in which gap matters.

Show the full solution

Seat the 6 boys first: 6! ways. This creates 7 gaps (one before, one after, and one between each pair of boys). Place the 4 girls into 4 of these 7 gaps, order mattering: 7P4 = 7x6x5x4 = 840. Total: 6! x 7P4 = 720 x 840 = 604800.

Answer: 604800

The trap

Using 7C4 instead of 7P4 for placing the girls loses the information about which specific girl sits in which gap — the girls are distinct people, so their arrangement within the chosen gaps matters too.

JEE AdvancedA recursive identity connecting consecutive combination termsAlgebraic identity proof

Prove that $^{n} C_{r} =^{n} C_{r - 1} \times \frac{n - r + 1}{r}$ for $1 \leq r \leq n$ .

Stuck? Show the approach

Write both combination terms out using the factorial formula and simplify their ratio directly — the identity falls out of cancelling shared factorial terms.

Show the full solution

$\frac{^{n} C _{r}}{^{n} C _{r - 1}} = \frac{n ! / [ r ! ( n - r )!]}{n ! / [( r - 1 )! ( n - r + 1 )!]} = \frac{( r - 1 )! ( n - r + 1 )!}{r ! ( n - r )!}$ . Since $r! = r \cdot (r - 1)!$ and $(n - r + 1)! = (n - r + 1) \cdot (n - r)!$ , this simplifies to $\frac{( n - r + 1 )}{r}$ . So $^{n} C_{r} =^{n} C_{r - 1} \times \frac{n - r + 1}{r}$ .

Answer: Proved by direct factorial simplification of the ratio

The trap

Trying to prove this via Pascal's identity instead of a direct ratio simplification leads to a much longer detour — this particular identity is a one-line cancellation once both terms are written in factorial form.

JEE MainCommittee selection with a minimum-count restrictionCase splitting

A committee of 5 is to be formed from 6 men and 5 women, including at least 3 men. Find the number of ways.

Stuck? Show the approach

List every valid split of the committee by the number of men (3, 4, or 5), compute each with combinations, then add, since the cases cannot occur simultaneously.

Show the full solution

3 men, 2 women: $^{6} C_{3} \times^{5} C_{2} = 20 \times 10 = 200$ . 4 men, 1 woman: $^{6} C_{4} \times^{5} C_{1} = 15 \times 5 = 75$ . 5 men, 0 women: $^{6} C_{5} \times^{5} C_{0} = 6 \times 1 = 6$ . Total: $200 + 75 + 6 = 281$ .

Answer: 281

The trap

Omitting the '5 men, 0 women' case because it feels like an edge case is a common oversight — 'at least 3 men' includes the possibility of an all-men committee whenever the group is large enough to fill it, which it is here (6 men available for 5 seats).

JEE MainDigit arrangement with a divisibility conditionMulti-case counting with a leading-digit restriction

How many 4-digit numbers divisible by 5 can be formed from the digits 0-9 without repeating any digit?

Stuck? Show the approach

A number is divisible by 5 exactly when its last digit is 0 or 5 — split into these two cases, and in the case ending in 5, separately subtract arrangements where the leading digit is 0.

Show the full solution

Case last digit = 0: the remaining 3 digits come from {1,...,9} (9 digits, no leading-zero risk since 0 is already used): $^{9} P_{3} = 504$ . Case last digit = 5: the remaining 3 digits come from the 9 digits excluding 5 (which include 0): $^{9} P_{3} = 504$ total ways to fill the first three positions, minus those with 0 leading, which is $^{8} P_{2} = 56$ (0 fixed first, 2 more digits from the remaining 8): $504 - 56 = 448$ . Total: $504 + 448 = 952$ .

Answer: 952

The trap

Forgetting to separately handle the leading-zero exclusion in the 'ends in 5' case (since 0 is available as one of the three remaining digits there, unlike the 'ends in 0' case where 0 is already used up) undercounts or overcounts depending on which shortcut is (wrongly) assumed.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 11 BoardHigh

JEE MainVery High

JEE AdvancedVery High

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Ask whether swapping the order of the chosen items changes the outcome. Seating people, forming a PIN, assigning distinct roles (President, VP) — order matters, use nPr. Choosing a team, a committee, a hand of cards — order doesn't matter, use nCr.

No. Circular permutations do not appear anywhere in the current 2026-27 NCERT chapter — not in the theorems, and not in any of its 42 exercise questions. It's commonly added by coaching material, but it isn't part of this specific book.

It's a definition chosen to make the formulas work: nP0 and nC0 should both equal 1, since there is exactly one way to arrange or select nothing at all. Setting 0!=1 is what makes n!/(n-0)! and n!/(0!n!) come out to 1 correctly.

Direct listing is usually far more complex than the complement (subtracting the 'together' count from the total) or the gap method (arranging everything else first, then placing the restricted objects into the resulting gaps) — both techniques turn a hard restriction into an easy one.

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