Class 11 Maths · Chapter 6
Short answer:
Class 11 Maths Chapter 6 — Permutations and Combinations counting ka chapter hai: kisi kaam ko kitne tareekon se kiya ja sakta hai, ye systematically nikalna sikhata hai. Do core ideas hain — permutation (order/arrangement matter karta hai) aur combination (sirf selection matter karta hai, order nahi). In dono ke saath Fundamental Principle of Counting (multiplication + addition principle) aur factorial notation is chapter ki neev hain.
Neeche is page par NCERT ke Exercise 6.1, 6.2 aur 6.3 ke saare 22 real questions step-by-step, poori working ke saath solve kiye gaye hain — agar tum poora class 11 maths ncert solutions series follow kar rahe ho, ye chapter Probability aur Binomial Theorem dono ki base banata hai.
Class 10 tak hum counting ke sawal ek-ek karke list bana ke solve kar lete the. Class 11 me numbers itne bade ho jaate hain (jaise "52 cards me se 5 card kitne tareeke se select honge") ki listing impossible ho jaati hai. Isliye Chapter 6 — Permutations and Combinations — counting ko ek formula-based science bana deta hai, bina ek-ek possibility likhe.
Poora chapter do sawalon ke around ghoomta hai: (1) kitne tareeke se arrange (order fix karke) kar sakte ho? — ye Permutation hai, aur (2) kitne tareeke se select (sirf group banake) kar sakte ho? — ye Combination hai. In dono ke beech ka fark hi is chapter ka sabse important concept hai, aur exam me sabse zyada marks isi confusion ki wajah se katte hain — committee/team select karni hai to nCr, password/arrangement banani hai to nPr.
Agar tumne class 11 maths ncert book pdf download kiya hai ya class 11 maths syllabus 2026-27 pdf check kiya hai, to dekhoge ye chapter Sets (class 11 maths chapter 1 sets ncert solutions se related) ke baad aata hai, aur aage Binomial Theorem, Sequences & Series (jiske class 11 maths chapter 8 sequences and series important questions alag se practice karne chahiye), Straight Lines (class 11 maths straight lines ncert solutions), Conic Sections (class 11 maths conic sections notes) aur Limits & Derivatives (class 11 maths limits and derivatives ncert solutions) tak connect hota rehta hai. Trigonometric Functions (class 11 maths chapter 3 trigonometric functions ncert solutions) is chapter se directly linked nahi hai, par same NCERT book ka part hai — sab ek saath revise karna best strategy hai.
Sabse zyada practical use — Chapter 13 Probability ka poora foundation Permutations & Combinations pe hi tika hai, isliye is chapter ko skip karna future me bahut mehenga padega.
Chapter 6 Summary — 5 Minute Revision
Chapter 6 me humne seekha: Fundamental Principle of Counting — agar ek kaam m tareeke se aur usse independent dusra kaam n tareeke se ho sakta hai, to dono successively m × n tareeke se; aur agar "ya to ye ya wo" (mutually exclusive) options hain to m + n tareeke se. Factorial (n!) repeated multiplication ko short-hand deta hai aur 0! = 1 define hota hai. Permutation (nPr = n!/(n−r)!) tab use hota hai jab r objects ko n distinct objects me se select karke ek particular order me arrange karna ho. Jab kuch objects repeat hote hain (jaise MISSISSIPPI ke letters), to permutation formula n!/(p1!·p2!·...) ban jaata hai. Combination (nCr = n!/[r!(n−r)!]) sirf selection (bina order ke) ke liye hai, aur nPr = nCr × r! is relation se dono jude hue hain.
In-text examples aur Exercises 6.1–6.3 ke saare 22 questions is page par step-by-step solve kiye gaye hain — number formation, word-letter arrangement (together/not together), committee selection, aur "at least/at most" wale mixed problems sabhi cover hue hain.
In-Text Questions — Solutions
Example 1 (Fundamental Principle — Multiplication). Ek restaurant ke menu me 3 tarah ke starters aur 4 tarah ke main course hain. Ek customer 1 starter aur 1 main course choose karta hai. Kitne alag-alag combinations ban sakte hain?
Starter choose karne ke 3 tareeke hain, aur har starter ke saath main course choose karne ke 4 tareeke hain (dono choices independent hain aur ek saath ho rahi hain — AND).
Total combinations = 3 × 4 = 12
Example 2 (Multiplication Principle — 3 stages). Delhi se Agra jaane ke 3 raaste hain aur Agra se Jaipur jaane ke 2 raaste hain. Delhi se Jaipur (via Agra) jaane ke kitne tareeke hain?
Delhi→Agra: 3 tareeke. Agra→Jaipur: 2 tareeke. Dono successive stages hain, isliye multiply karenge.
Total = 3 × 2 = 6
Example 3 (Addition Principle — OR). Ek club me 3 male aur 4 female senior members hain. Club ka ek representative (koi bhi 1 senior member) choose karna hai. Kitne tareeke se ye ho sakta hai?
Ya to ek male choose hoga (3 tareeke) ya ek female (4 tareeke) — ye do options mutually exclusive hain (ek member ya male hai ya female, dono nahi), isliye add karenge, multiply nahi.
Total = 3 + 4 = 7
Example 4 (Factorial). 5!/3! ka value nikalo, aur batao 0! = 1 kyun define kiya jaata hai.
5! = 5×4×3×2×1 = 120, aur 3! = 3×2×1 = 6.
5!/3! = 120/6 = 20 (shortcut: 5!/3! = 5×4 = 20)
0! = 1 isliye define hota hai taaki nPr aur nCr ke formulas r = n wale case me bhi consistent rahein (jaise n!/(n−n)! = n!/0! = n!/1, jo sahi answer deta hai).
Example 5 (Direct nPr application). Ek shelf par 5 alag-alag books hain, jisme se 3 books choose karke unhe ek line me arrange karna hai. Kitne tareeke se ho sakta hai?
Yahan order matter karta hai (arrangement), isliye permutation formula use karenge: n = 5, r = 3.
5P3 = 5!/(5−3)! = 5!/2! = (5×4×3×2!)/2! = 5×4×3 = 60
Example 6 (Permutation with repetition allowed). Ek 4-digit PIN banana hai using digits 0–9, jisme digits repeat ho sakte hain. Kitne PIN possible hain?
Har ek digit-position ke liye independently 10 choices hain (0 se 9), aur repetition allowed hai, isliye multiplication principle seedha lagta hai (permutation formula ki zaroorat nahi kyunki repetition allowed hai).
Total PINs = 10 × 10 × 10 × 10 = 10⁴ = 10,000
Example 7 (All distinct objects, all at a time). 6 dost ek row me kitne tareeke se baith sakte hain?
6 distinct persons, sabko ek row (6 positions) me arrange karna hai — ye 6P6 = 6! hai.
6! = 6×5×4×3×2×1 = 720
Example 8 (Restricted permutation — always together). 5 students ek row me baithne hain, par 2 particular students (A aur B) hamesha saath (adjacent) baithne chahiye. Kitne tareeke se ye ho sakta hai?
A aur B ko ek single unit (block) maan lo. Ab total units = 4 (3 baaki students + 1 AB-block), jinhe arrange karne ke 4! tareeke hain. Block ke andar A aur B do tareeke se arrange ho sakte hain (AB ya BA), isliye 2!.
Total = 4! × 2! = 24 × 2 = 48
Example 9 (Restricted permutation — never together). Same 5 students (Example 8 wale) ko row me baithana hai par A aur B kabhi saath nahi baithne chahiye. Kitne tareeke se ho sakta hai?
Total arrangements (bina kisi restriction ke) − arrangements jisme A,B saath hain (Example 8 se).
Total unrestricted = 5! = 120
A,B kabhi saath nahi = 120 − 48 = 72
Example 10 (Permutations of objects not all distinct). Word 'INDIA' ke letters se kitne distinct arrangements (words) banaye ja sakte hain?
INDIA me 5 letters hain: I, N, D, I, A — yahan 'I' 2 baar repeat ho raha hai, baaki (N, D, A) distinct hain.
Arrangements = 5!/2! = 120/2 = 60
Example 11 (Basic Combination). Ek basket me 5 alag-alag fruits hain. Inme se koi bhi 2 fruits choose karne hain (order matter nahi karta). Kitne tareeke se ye ho sakta hai?
Yahan sirf selection karni hai, arrangement nahi — isliye combination formula: n = 5, r = 2.
5C2 = 5!/(2!×3!) = (5×4)/(2×1) = 10
Example 12 (Relation between nPr and nCr). Verify karo ki 6P3 = 6C3 × 3!
6P3 = 6!/(6−3)! = 6!/3! = 720/6 = 120
6C3 = 6!/(3!×3!) = 720/(6×6) = 20
6C3 × 3! = 20 × 6 = 120 = 6P3 ✓ (Verified — combination pehle select karti hai, phir permutation us selection ko arrange karti hai)
Example 13 (Combination — diagonals application). Ek hexagon (6 vertices) ke kitne diagonals hote hain?
Pehle 6 vertices me se koi bhi 2 choose karke total line segments nikalte hain: 6C2 = 15. In 15 me se 6 line segments hexagon ke sides hain (diagonals nahi). Baaki diagonals hain.
Diagonals = 6C2 − 6 = 15 − 6 = 9
Example 14 (Combination — committee with a condition). 6 boys aur 4 girls me se 4-member committee banani hai jisme kam se kam 1 girl zaroor ho. Kitne tareeke se ho sakta hai?
Direct method (complement) easier hai: Total committees (bina restriction) − committees jisme koi girl nahi (sirf boys).
Total = 10C4 = 210
Koi girl nahi (sirf boys se) = 6C4 = 15
Kam se kam 1 girl = 210 − 15 = 195
Poore Class 11 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q22)
Exercise 6.1, Q1. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that (i) repetition of the digits is allowed? (ii) repetition of the digits is not allowed?
(i) Repetition allowed: har ek 3 positions ke liye independently 5 choices hain.
Total = 5 × 5 × 5 = 125
(ii) Repetition not allowed: pehla digit 5 tareeke se, dusra baaki 4 me se, teesra baaki 3 me se.
Total = 5 × 4 × 3 = 60
Exercise 6.1, Q2. How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?
Ek number even tabhi hoga jab uska units (last) digit even ho. Digits {1,2,3,4,5,6} me se even digits: 2, 4, 6 — 3 choices. Hundreds aur tens position par repetition allowed hai, isliye 6-6 choices.
Total = 6 × 6 × 3 = 108
Exercise 6.1, Q3. How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?
10 letters me se 4 positions fill karni hain, repetition nahi.
Total = 10 × 9 × 8 × 7 = 5040
Exercise 6.1, Q4. How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?
Pehle 2 digits fixed hain (6, 7). Baaki 3 digits, remaining 8 digits (0-9 me se 6,7 nikaal ke) me se, bina repetition ke, choose karne hain.
Total = 8 × 7 × 6 = 336
Exercise 6.1, Q5. A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?
Har toss ke 2 possible outcomes hain (Head/Tail), 3 tosses independent hain.
Total = 2 × 2 × 2 = 8
Exercise 6.1, Q6. Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
Yahan order matter karta hai (upar wala flag alag signal, neeche wala alag), aur repetition nahi (same flag dono positions me use nahi hoga).
Total = 5 × 4 = 20
Exercise 6.2, Q1. Evaluate (i) 8! (ii) 4! − 3!
(i) 8! = 1×2×3×4×5×6×7×8
8! = 40320
(ii) 4! = 24, 3! = 6
4! − 3! = 24 − 6 = 18
Exercise 6.2, Q2. Is 3! + 4! = 7!?
3! = 6, 4! = 24, isliye 3! + 4! = 6 + 24 = 30. Aur 7! = 5040.
30 ≠ 5040 ⇒ 3! + 4! ≠ 7!
Isliye statement galat hai (No).
Exercise 6.2, Q3. Compute 8!/(6! × 2!)
8! = 8×7×6!, isliye 6! cancel ho jaayega.
8!/(6!×2!) = (8×7×6!)/(6!×2!) = (8×7)/2! = 56/2 = 28
Exercise 6.2, Q4. If 1/6! + 1/7! = x/8!, find x.
Dono side ko 8! se multiply karo:
8!/6! + 8!/7! = x
8!/6! = 8×7 = 56, aur 8!/7! = 8
x = 56 + 8 = 64
Exercise 6.2, Q5. Evaluate n!/(n−r)! when (i) n = 6, r = 2 (ii) n = 9, r = 5
(i) 6!/(6−2)! = 6!/4! = (6×5×4!)/4! = 6×5 = 30
(ii) 9!/(9−5)! = 9!/4! = 9×8×7×6×5 = 15120
Exercise 6.3, Q1. How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
9 digits available, 3 positions, bina repetition ke — ye seedha 9P3 hai.
9P3 = 9 × 8 × 7 = 504
Exercise 6.3, Q2. How many 4-digit numbers are there with no digit repeated?
4-digit number ka pehla digit 0 nahi ho sakta. Pehla digit: 1-9 me se koi bhi = 9 choices. Baaki 3 digits remaining 9 digits (0 included, jo already use hue digit chhod ke) me se bina repetition ke fill honge.
Total = 9 × 9 × 8 × 7 = 4536
Exercise 6.3, Q3. How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7 if no digit is repeated?
Units digit even hona chahiye: {2, 4, 6} me se — 3 choices. Baaki 2 positions (hundreds, tens) remaining 5 digits me se bina repetition ke fill honge.
Units digit fix hone ke baad baaki 2 places = 5 × 4 = 20
Total = 3 × 20 = 60
Exercise 6.3, Q4. Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?
Total 4-digit numbers: 5 digits me se 4 positions, bina repetition.
5P4 = 5 × 4 × 3 × 2 = 120
Even numbers ke liye units digit {2, 4} me se hona chahiye — 2 choices. Baaki 3 positions remaining 4 digits me se bina repetition ke.
Even numbers = 2 × (4 × 3 × 2) = 2 × 24 = 48
Exercise 6.3, Q5. From a committee of 8 persons, in how many ways can we choose a chairman and a vice-chairman assuming one person cannot hold more than one position?
Chairman: 8 tareeke se choose. Vice-chairman: baaki bache 7 me se koi bhi.
Total = 8 × 7 = 56
Exercise 6.3, Q6. Find n if n−1P3 : nP4 = 1 : 9.
n−1P3 = (n−1)!/(n−4)!, aur nP4 = n!/(n−4)!
(n−1)!/(n−4)! ÷ n!/(n−4)! = (n−1)!/n! = 1/n
Ratio diya hai 1/9, isliye:
1/n = 1/9 ⇒ n = 9
Exercise 6.3, Q7. Find r if (i) 5Pr = 2 6Pr−1 (ii) 5Pr = 6Pr−1
(i) 5Pr = 5!/(5−r)! aur 6Pr−1 = 6!/(7−r)!
120/(5−r)! = 2 × 720/(7−r)!
(7−r)!/(5−r)! = 1440/120 = 12
(7−r)(6−r) = 12 ⇒ r² − 13r + 30 = 0
r = 10 ya r = 3
Kyunki 5Pr sirf r ≤ 5 ke liye defined hai, isliye r = 3.
(ii) 5Pr = 6Pr−1
120/(5−r)! = 720/(7−r)!
(7−r)(6−r) = 6 ⇒ r² − 13r + 36 = 0
r = 9 ya r = 4
r ≤ 5 hona chahiye, isliye r = 4.
Exercise 6.3, Q8. How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?
EQUATION me 8 letters hain: E, Q, U, A, T, I, O, N — sabhi distinct hain.
Total words = 8! = 40320
Exercise 6.3, Q9. How many words, with or without meaning, can be made from the letters of the word MONDAY, assuming that no letter is repeated, if (i) 4 letters are used at a time, (ii) all letters are used at a time, (iii) all letters are used but first letter is a vowel?
MONDAY me 6 distinct letters hain: M, O, N, D, A, Y.
(i) 4 letters at a time:
6P4 = 6 × 5 × 4 × 3 = 360
(ii) Saare 6 letters:
6P6 = 6! = 720
(iii) Pehla letter vowel ho: MONDAY ke vowels — O, A (2 vowels). Pehli position: 2 choices. Baaki 5 letters, 5 positions me arrange: 5!.
Total = 2 × 5! = 2 × 120 = 240
Exercise 6.3, Q10. In how many of the distinct permutations of the letters in MISSISSIPPI do the four I's not come together?
MISSISSIPPI me 11 letters hain: M(1), I(4), S(4), P(2).
Total distinct permutations = 11!/(4!×4!×2!) = 34650
Ab 4 I's ko ek block maan lo. Units: M, S,S,S,S, P,P, (IIII) → total 8 units, jisme S 4 baar aur P 2 baar repeat.
I's together = 8!/(4!×2!) = 40320/48 = 840
I's not together = 34650 − 840 = 33810
Exercise 6.3, Q11. In how many ways can the letters of the word PERMUTATIONS be arranged if the (i) words start with P and end with S, (ii) vowels are all together, (iii) there are always 4 letters between P and S?
PERMUTATIONS me 12 letters hain, jisme T do baar repeat hota hai (P,E,R,M,U,T,A,T,I,O,N,S); baaki sab distinct.
(i) P fix pehle position, S fix last position. Baaki 10 letters (jisme T 2 baar repeat) beech ki 10 positions me arrange:
10!/2! = 3628800/2 = 1814400
(ii) Vowels: E, U, A, I, O (5, sab distinct). Consonants: P, R, M, T, T, N, S (7, T repeat 2 baar). Vowels ko ek block maan lo → total 8 units (7 consonants + 1 vowel-block).
8 units arrange = 8!/2! = 20160
Vowel block ke andar arrange = 5! = 120
Total = 20160 × 120 = 2419200
Important Equations — Ek Nazar Me
| Concept | Formula / Rule |
|---|---|
| Multiplication Principle (AND) | Agar kaam A, m tareeke se aur kaam B, n tareeke se ho sakta hai (independent, successive), to dono saath = m × n tareeke |
| Addition Principle (OR) | Agar kaam A, m tareeke se ya kaam B, n tareeke se (mutually exclusive) ho sakta hai, to = m + n tareeke |
| Factorial | n! = n × (n−1) × (n−2) × ... × 2 × 1 | 0! = 1 |
| Permutation | nPr = n!/(n−r)! , 0 ≤ r ≤ n |
| Arrangements with repetition allowed (r positions, n choices each) | nr |
| All n distinct objects taken all at a time | n! |
| Permutations when objects are not all distinct | n!/(p1! × p2! × ... × pk!) — jahan p1, p2... same-type objects ki repetition count hai |
| Combination | nCr = n! / [r! (n−r)!] , 0 ≤ r ≤ n |
| Relation between nPr aur nCr | nPr = nCr × r! |
| Symmetry property | nCr = nC(n−r) |
| Pascal-type identity | nCr + nC(r−1) = (n+1)Cr |
| nC0 aur nCn | nC0 = nCn = 1 |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Permutation aur combination me confuse ho jaana — jab order/arrangement matter karta hai (password, seating, ranking) wahan bhi galti se ⁿCᵣ laga dena, ya jab sirf selection/group banani hai wahan ⁿPᵣ laga dena.
- "n-digit number" wale sawal me first digit ko 0 allow kar dena — leading digit kabhi 0 nahi ho sakta, isliye first place ke liye choices ek kam ho jaati hain (jaise Ex 6.3 Q2 me).
- Word ke letters arrange karte waqt repeated letters (jaise MISSISSIPPI ka I, S; INDIA ka I) ko distinct maan lena aur seedha n! laga dena — jabki repeated objects ke liye n!/(p₁!×p₂!...) se divide karna zaroori hai.
- "Sab saath (together)" wale restricted-permutation sawal me sirf block ko ek unit maan ke bahar arrange kar dena, par block ke andar internal arrangement (r!) multiply karna bhool jaana.
- ⁿ⁻¹P₃ : ⁿP₄ jaisi algebraic equations me factorial ratio (jaise (n−1)!/n! = 1/n) simplify karte waqt sign ya cancellation ki galti kar dena, jisse final n ya r ka wrong value aa jaata hai.
- "At least" / "at most" wale combination sawalon me sab cases (0,1,2,3... girls) me se kuch case chhod dena, ya disjoint cases ko multiply karna jabki unhe add (OR) karna chahiye tha.
Board-Style Important Questions
- Practice Question (Very Short Answer): Evaluate 0! + 1!.
Solution: 0! = 1, 1! = 1, isliye 0! + 1! = 1 + 1 = 2. - Practice Question (Short Answer): If nC9 = nC8, find nC17.
Solution: nCr = nCs tab hota hai jab r = s ya r + s = n. Yahan r ≠ s hai, isliye 9 + 8 = n ⇒ n = 17. Ab 17C17 = 1. - Practice Question (Short Answer): In how many ways can 4 boys and 3 girls be seated in a row so that no two girls sit together?
Solution: Pehle 4 boys ko row me arrange karo: 4! = 24 tareeke — isse 5 gaps ban jaate hain (_B_B_B_B_). In 5 gaps me se 3 gaps choose karke 3 girls ko arrange karo: 5P3 = 5×4×3 = 60. Total = 24 × 60 = 1440. - Practice Question (Long Answer): How many different words (with or without meaning) can be formed using all the letters of the word ALLAHABAD?
Solution: ALLAHABAD me 9 letters hain: A (4 baar), L (2 baar), H, B, D (1-1 baar). Total arrangements = 9!/(4!×2!) = 362880/48 = 7560. - Practice Question (Long Answer): A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has at least 3 girls?
Solution: Case (exactly 3 girls, 2 boys): 4C3 × 7C2 = 4 × 21 = 84. Case (exactly 4 girls, 1 boy): 4C4 × 7C1 = 1 × 7 = 7. Dono disjoint cases hain, isliye add karo: 84 + 7 = 91.
Aksar Poochhe Jaane Wale Sawaal
Class 11 Maths Chapter 6 Permutations and Combinations me kitne exercises hain aur total kitne questions hain?
Poore chapter me 4 exercises (6.1, 6.2, 6.3, 6.4) aur 1 Miscellaneous Exercise hain. Is page par abhi Exercise 6.1, 6.2 aur 6.3 ke saare 22 questions step-by-step solve kiye gaye hain — Exercise 6.4 aur Miscellaneous Exercise jald hi add honge.
Permutation aur Combination me basic fark kya hai?
Permutation me order/arrangement matter karta hai (jaise seating, password, ranking) — formula nPr = n!/(n−r)!. Combination me sirf selection/grouping matter karta hai, order nahi (jaise committee, team) — formula nCr = n!/[r!(n−r)!]. Relation: nPr = nCr × r!.
Is chapter ka Probability ke saath kya connection hai?
Chapter 13 Probability me sample space ke outcomes count karne, ya kisi event ke favourable outcomes nikalne ke liye seedha Permutations & Combinations use hote hain. Isliye class 11 maths probability important questions practice karne se pehle ye chapter clear hona zaroori hai.
Is chapter ke formulas ek jagah PDF me kahan milenge?
Upar diya gaya formulas table hi is chapter ke saare key formulas (n!, nPr, nCr, aur identities) cover karta hai — isse save/print karke apna khud ka class 11 maths formulas pdf download jaisa quick-revision sheet ban sakta hai.
NCERT Exemplar se is chapter ke extra questions practice karne chahiye?
Haan — NCERT textbook ke questions basics clear karte hain, par HOTS aur tricky "at least/at most" wale word problems ke liye class 11 maths exemplar solutions zaroor practice karo, especially restricted-arrangement aur mixed permutation-combination wale sawal.
2026-27 session ke rationalised syllabus me is chapter ka koi topic drop hua hai?
Permutations and Combinations chapter khud fully intact hai. Poore Class 11 Maths book me rationalisation ke through jo do poore chapters drop hue — Principle of Mathematical Induction aur Mathematical Reasoning — unka is chapter se koi lena-dena nahi hai. Exact current list ke liye class 11 maths syllabus 2026-27 pdf ncert.nic.in se check kar sakte ho.
Class 11 Maths — Saare Chapters
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