NCERT Plus One (Class 11) Mathematics • Chapter 6

Permutations and Combinations

Fundamental counting principle, factorials, nPr arrangements (with repeats) & nCr selections.

📄 Official NCERT Textbook PDF

Topic 6.2: Fundamental Principle of Counting (Multiplication & Addition)

Getting dressed for Onam at school, you open your cupboard and spot 3 shirts (Blue, Cream, Green) and 2 mundus (White, Kasavu). Each shirt pairs with either mundu, giving 3 × 2 = 6 outfits without listing them all! Now suppose at the canteen you want EITHER 1 of 4 dosas OR 1 of 3 appams (just one snack!) — you pick 4 + 3 = 7 ways. Multiplication for AND, Addition for OR!
MULTIPLICATION (AND)Step 1 in m ways AND Step 2 in n ways → m × n total ways
ADDITION (OR)Task A in m ways OR disjoint Task B in n ways → m + n ways
SLOT BOX METHODDraw 1 box per choice, fill restricted box FIRST, then multiply!
🧠 Memory Trick: AND = Multiply (×) when events happen one after another; OR = Add (+) when you pick only one option!
💡 Simple Kid-Check: Imagine a branching road map: 3 roads lead from Kochi to Thrissur, and 2 roads lead from Thrissur to Palakkad. Each of the 3 first roads splits into 2 second roads, creating 3 × 2 = 6 full journeys!
  • 1
    Draw Empty Boxes: Forming a 3-digit PIN? Draw [ _ ] [ _ ] [ _ ].
  • 2
    Fill Conditions First: Even number? Zero forbidden at start? Always lock that restricted box first!
  • 3
    Check Repetition Rule: "Digits repeated" keeps counts equal; "distinct / no repeat" drops count by 1 per box.
🧒 Super-Easy Example: How many 3-digit EVEN numbers can be formed from {1, 2, 3, 4, 5, 6} if digits cannot be repeated?
• Fill the units box FIRST (must be even: 2, 4, 6) → 3 choices.
• Now 5 digits remain for the hundreds box → 5 choices.
• Then 4 digits remain for the tens box → 4 choices.
• Total = 5 × 4 × 3 = 60 numbers! (With repetition: 6 × 6 × 3 = 108.)
Watch TWO classic exam traps: (1) Filling left-to-right when the units digit is restricted (like "even" or "divisible by 5") — always fill the restricted spot FIRST! (2) When 0 is in the digit set, 0 can NEVER sit in the first (leftmost) digit of a 3-digit or 4-digit number!

Topic 6.3: Factorial Notation (n! & 0! = 1) & Simple Simplifications

Four friends — Anu, Babu, Chitra, and Dev — line up for a selfie in front of Kozhikode beach. Who stands first? 4 choices! Next spot? 3 left, then 2, then 1: 4 × 3 × 2 × 1 = 24! Mathematicians got tired of writing long countdown products, so they invented the excited exclamation mark n! ("n factorial") to shrink 4 × 3 × 2 × 1 into sleek 4!!
DEFINITIONn! = n × (n − 1) × (n − 2) × ... × 2 × 1 (for natural n)
ZERO FACTORIAL0! = 1 (by definition — 1 way to arrange an empty set!)
PEEL-OFF RULEn! = n × (n − 1)! = n(n − 1)(n − 2)! (cancel big fractions fast!)
🧠 Memory Trick: Think of n! like Russian nested dolls — peel off n, and (n − 1)! waits neatly inside to cancel denominators!
💡 Simple Kid-Check: Factorials explode super fast! Notice how 5! = 120 is just 5 friends in a queue, while 10! = 3,628,800 is over three and a half million queues! Never calculate giant numbers first — ALWAYS peel and cancel!
n01234567
n!1126241207205040
Peel-Off Trick: 8! = 8 × 7 × 6!  |  So 8!6! · 2! = 8 × 7 × 6!6! × 2 = 28!
🧒 Classic DHSE Equation: If 16! + 17! = x8!, find x.
Peel everything to match 6!: 16! + 17 · 6! = x8 · 7 · 6!.
Cancel 1/6! everywhere: 1 + 17 = x56 ⇒ 87 = x56 ⇒ x = 64!
Factorials DO NOT distribute over +, −, or ÷! (3 + 2)! = 120 is NOT 3! + 2! = 8, and 6!2! = 360 is NOT 3! = 6! Also remember 0! = 1, NOT 0 — otherwise formulas dividing by 0! would blow up!

Topic 6.3.1: Permutations of Distinct Objects: nPr = n! / (n − r)!

At your school sports day, 8 sprinters race in the 100 m final, and there are 3 podium medals: Gold, Silver, and Bronze. Picking (Asha = Gold, Binu = Silver, Cinu = Bronze) is totally different from (Binu = Gold, Asha = Silver, Cinu = Bronze)! When ORDER / RANK MATTERS, every arrangement is called a Permutation!
PERMUTATION FORMULAⁿPᵣ = n!(n − r)! = n(n−1)...(n−r+1) (0 ≤ r ≤ n)
BOUNDARY VALUESⁿPₙ = n! (all objects)  |  ⁿP₀ = 1  |  ⁿP₁ = n
WITH REPETITIONIf an object can repeat in every slot → nʳ (NOT ⁿPᵣ!)
🧠 Memory Trick: P = Position Matters! To compute ⁷P₃ fast without factorials, just multiply 3 steps down from 7: 7 × 6 × 5 = 210!
💡 Simple Kid-Check: Why does ⁿPᵣ divide n! by (n − r)!? Because you only fill r chairs out of n people — dividing by (n − r)! cleanly chops off the tail of people who didn't get a chair!
  • 1
    Distinct (No Repeats): Filling r slots from n distinct items = n(n − 1)(n − 2)... (r factors) = ⁿPᵣ.
  • 2
    Repetition Allowed: Filling r slots when you can reuse any of the n items = n × n × ... × n = nʳ.
🧒 Classic Equation Solve: Find n if ⁿP₅ = 42 · ⁿP₃ (with n > 4).
Expand descending steps: n(n − 1)(n − 2)(n − 3)(n − 4) = 42 · n(n − 1)(n − 2).
Since n > 4, cancel n(n − 1)(n − 2) from both sides:
(n − 3)(n − 4) = 42 = 7 × 6! Matching the bigger factors directly gives n − 3 = 7 ⇒ n = 10! (No quadratic formula needed!)
Watch TWO traps in ⁿPᵣ equations: (1) Always verify the domain n ≥ r! When solving (n − 3)(n − 4) = 42 algebraically gives n = 10 or n = −3, you MUST reject −3 because n must be a natural number ≥ 5! (2) Watch words like "repetition allowed" — that instantly switches from ⁿPᵣ to nʳ!

Topic 6.3.4: Permutations with Identical Letters & String Conditions

Look at the word MOM. If the two identical Ms trade places, the word still reads MOM! You didn't create a new arrangement — you created a twin clone! So whenever letters or objects repeat, divide by the factorial of each twin group. And if vowels insist on sitting next to each other on a school bus? Tie them inside one Vowel Bag!
IDENTICAL OBJECTSOut of n items, if p₁, p₂, ... repeat → n!p₁! p₂! ...
TOGETHER ("VOWEL BAG")Tie glued items as 1 unit, arrange outer units, × internal shuffles!
NONE TOGETHER (3+ ITEMS)Use GAP METHOD! Arrange others first → place items in gaps!
🧠 Memory Trick: For 2 items never together, Total − Together works. For 3 or more items never together, ALWAYS use the GAP METHOD!
💡 Simple Kid-Check: Imagine putting 4 identical blue marbles and 2 identical red marbles in a row. Swapping two blue marbles changes nothing! Dividing by 4! · 2! erases all invisible twin swaps in one clean stroke!
🏆 Super-Famous NCERT Word: MISSISSIPPI (11 letters total)
Count twins: M = 1, I = 4, S = 4, P = 2.
• Total permutations: 11!4! · 4! · 2! = 34,650.
• All 4 I's DO NOT come together: First glue [IIII] as 1 super-letter → 8 units left (with 4 S, 2 P). Inside [IIII], 4 identical I's arrange in 4!/4! = 1 way! So "4 I's together" = 8!4! · 2! = 840. Subtract: 34,650 − 840 = 33,810!
🔤 Distinct Word with Vowel Bag: DAUGHTER (8 distinct letters; vowels A, U, E)
• Vowels together: Glue [AUE] + 5 consonants = 6 units → 6! × 3! = 720 × 6 = 4,320.
• No two vowels together (Gap Method): Arrange 5 consonants in 5! = 120 ways, creating 6 gaps _ C _ C _ C _ C _ C _. Pick & arrange 3 vowels in 6 gaps: ⁶P₃ = 120. Total = 120 × 120 = 14,400!
HUGE DHSE Trap: In DAUGHTER, "no two vowels together" is NOT 8! − (vowels together)! Why? Because 8! − 4320 = 36,000 only removes cases where ALL 3 vowels sit together — leaving cases where 2 vowels still sit next to each other! For "NO TWO together" among 3+ items, ALWAYS use the Gap Method!

Topic 6.4: Combinations: nCr = n! / (r!(n − r)!) (Order Doesn't Matter)

Imagine picking 3 classmates from a group of 10 friends to go buy samosas from the canteen! Whether you call out Anu, Manu, Sanu or Sanu, Manu, Anu, it is the EXACT SAME team of 3 walking to the counter! When you only care WHO is picked (and NOT the order), it is a Combination (ⁿCᵣ)!
COMBINATION FORMULAⁿCᵣ = n!r!(n − r)! = ⁿPᵣr! (Order ignored!)
SYMMETRY RULEⁿCᵣ = ⁿCₙ₋ᵣ (e.g. ¹⁰⁰C₉₈ = ¹⁰⁰C₂ = 4,950!)
EQUALITY PROPERTYIf ⁿCₐ = ⁿC_b, then either a = b OR a + b = n!
🧠 Memory Trick: C = Choose / Committee! Selecting 3 friends to come WITH you = selecting (n − 3) friends to stay BEHIND — that's why ⁿCᵣ = ⁿCₙ₋ᵣ!
💡 Simple Kid-Check: Why is ⁿCᵣ = ⁿPᵣ / r!? Every group of r chosen friends can scramble into r! different queues! Dividing by r! collapses all r! queues of the same group down to 1 single selection!
Must-Know Shortcut Values: ⁿC₀ = 1  |  ⁿC₁ = n  |  ⁿCₙ = 1  |  ⁿC₂ = n(n−1)2
🧒 Instant 2-Mark Board Question: If ⁿC₈ = ⁿC₂, find ⁿC₂.
Since 8 ≠ 2, use the complementary rule: n = 8 + 2 = 10.
Now compute ¹⁰C₂ = 10 × 92 × 1 = 45!
Watch TWO common blunders: (1) Thinking ⁿCₐ = ⁿC_b only implies a = b — always check the complementary branch a + b = n! (2) Confusing Selection (ⁿCᵣ) with Arrangement (ⁿPᵣ): choosing a Captain + Vice-Captain has different roles so it is ⁿP₂, whereas choosing 2 co-captains with identical roles is ⁿC₂!

Topic 6.4: Pascal's Identity & Geometric / Committee Selection Problems

How many handshakes happen if 10 guests shake hands once at a Thrissur wedding? Every handshake picks 2 people: ¹⁰C₂ = 45! How many triangles from 10 dots on a circle? Pick 3 dots: ¹⁰C₃ = 120! What if you need a school quiz team with "at least 1 girl"? Split cleanly into exact cases — or use Total − No Girls!
PASCAL'S IDENTITYⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ (Same top n, adjacent bottoms!)
GEOMETRY COUNTSChords = ⁿC₂  |  Diagonals = ⁿC₂ − n  |  Triangles = ⁿC₃ − ᵏC₃ (k collinear)
"AT LEAST" RULEMultiply within a case (AND), add across cases (OR) — or use Total − Excluded!
🧠 Memory Trick: For Pascal's Identity, if the two upper numbers match (n) and lower numbers are neighbours (r and r−1), bump the top up to n+1 and keep the BIGGER bottom r!
💡 Simple Kid-Check: Why are diagonals of an n-gon equal to ⁿC₂ − n? Joining any 2 vertices creates ⁿC₂ lines total — n of those are outer polygon sides, so subtracting n leaves only interior diagonals!
🏆 Classic 4-Mark Committee Problem: From 4 girls and 7 boys, form a team of 5 with at least 3 girls:
Split into mutually exclusive exact cases:
• Case 1 (3 girls AND 2 boys): ⁴C₃ × ⁷C₂ = 4 × 21 = 84
• Case 2 (4 girls AND 1 boy): ⁴C₄ × ⁷C₁ = 1 × 7 = 7
Add both cases (OR): 84 + 7 = 91 teams!
🃏 Classic Deck Selection: How many 5-card hands from 52 cards have exactly 1 Ace?
Pick 1 Ace out of 4 Aces AND 4 non-Aces out of remaining 48 cards = ⁴C₁ × ⁴⁸C₄!
NEVER solve "at least 1 girl" by picking ⁴C₁ first and filling remaining 4 slots from everyone else (¹⁰C₄)! Why? Because ⁴C₁ × ¹⁰C₄ double-counts every team with 2+ girls! ALWAYS list disjoint exact-count cases (1G+4B, 2G+3B, ...) OR use the lightning-fast complement trick: Total − All Boys!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 1 Marks): Evaluate 7! / 5!

Key (1 Mark): Peel down 7!:
7 × 6 × 5!5! = 7 × 6 = 42.

Q2 (Kerala HSE Say • 1 Marks): How many chords can be drawn through 21 points on a circle?

Each chord joins 2 points (order doesn't matter):
²¹C₂ = 21 × 202 × 1 = 210 chords.

Q3 (Kerala HSE March • 2 Marks): How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that (i) repetition is allowed, (ii) repetition is not allowed?

• (i) Repetition allowed: 5 × 5 × 5 = 125.
• (ii) Repetition not allowed: 5 × 4 × 3 = 60.

Q4 (NCERT / DHSE Model • 2 Marks): Find n if ⁿ⁻¹P₃ : ⁿP₄ = 1 : 9.

(n−1)! / (n−4)!n! / (n−4)! = 19
Cancel (n−4)! and peel n! = n(n−1)!:
1n = 19 ⇒ n = 9.

Q5 (Kerala HSE March • 3 Marks): Find the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that (i) all vowels occur together, (ii) all vowels do not occur together.

8 distinct letters; 3 vowels (A, U, E), 5 consonants.
• (i) Glue [AUE] as 1 unit → 6 units: 6! × 3! = 720 × 6 = 4,320.
• (ii) All vowels NOT together = Total − Together = 8! − 4,320 = 40,320 − 4,320 = 36,000.

Q6 (Kerala HSE Say • 3 Marks): In how many ways can the letters of the word ASSASSINATION be arranged so that all the S's are together?

13 letters: A(3), S(4), I(2), N(2), T(1), O(1).
Tie 4 S's as 1 block [SSSS] → 10 units total.
Internal S arrangements = 4!/4! = 1.
Total = 10!3! · 2! · 2! = 3,628,80024 = 1,51,200.

Q7 (Kerala HSE March • 3 Marks): A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.

Select 2 black from 5 AND 3 red from 6:
⁵C₂ × ⁶C₃ = 5×42 × 6×5×46 = 10 × 20 = 200 ways.

Q8 (NCERT / Kerala HSE • 4 Marks): 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 (i) no girl, (ii) at least one boy and one girl, (iii) at least 3 girls?

• (i) No girl (all 5 boys): ⁷C₅ = ⁷C₂ = 21.
• (ii) At least 1 boy & 1 girl: Total − (No girl) − (No boy) = ¹¹C₅ − ⁷C₅ − 0 = 462 − 21 = 441.
• (iii) At least 3 girls: ⁴C₃×⁷C₂ + ⁴C₄×⁷C₁ = 4(21) + 1(7) = 84 + 7 = 91.