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State Board · Class 11 · Mathematics Part 2

Lesson 3: Permutations and Combinations

5 concepts · 40 practice questions with answers · Free to read

This lesson is also in: मराठी

Fundamental principles of counting

From daily life

When choosing a shirt and a trouser each day we count the options in our head. The principles of counting give the number quickly without writing the list. Building on classes 9–10 : Since classes 9–10 we have counted numbers and sets; here we make it systematic with the multiplication and addition principles.

Simple definition

If two tasks are done one after the other in m and n ways, both together can be done in m × n ways (multiplication principle). If only one of two tasks is to be done, it can be done in m + n ways (addition principle).

Rule 1: When to multiply, when to add — "And", "then", "both" mean multiply. "Or", "only one" mean add. When forming numbers, fill the place that has a condition first.

S1T1T2S2T1T2S3T1T26outfits3 × 2 = 6 (multiplication principle)

In the figure : Tree diagram: 3 shirts × 2 trousers with each = 6 outfits.

How is the formula obtained? : Each time the first task is done in one of m ways, the second task can be done in n ways; that is m groups of n results, m × n in all. In the figure there are 3 branches with 2 ends each.

  • Easy : 3 shirts and 4 trousers: 3 × 4 = 12 outfits.
  • Medium : Three-digit numbers from 1, 2, 3, 4, 5 without repetition: 5 × 4 × 3 = 60.
  • Exam-type : Three-digit even numbers: hundreds 9, tens 10, units 5 (0, 2, 4, 6, 8); 9 × 10 × 5 = 450.
Remember

Fill the place with the condition first: the units place for an even number, the first place when 0 is not allowed.

Common mistake

Taking 4-digit numbers as 10⁴. The first digit cannot be 0, so 9 × 10 × 10 × 10 = 9000.

Shortcut

For "at least one" take total − none; know the counts: 2-digit 90, 3-digit 900, 4-digit 9000. Memory clue : "Nine, then a row of zeros".

CET trap

When all digits must differ the first is 9 ways, the second 9 ways (0 is allowed), then 8, 7; do not take 8 for the second.

CET-style example : The number of 4-digit numbers with all digits different? (A) 4536 (B) 5040 (C) 9000 (D) 3024

Answer : (A). 9 × 9 × 8 × 7 = 4536. Time-saving method : First place 9, then 9, 8, 7; 5040 is ¹⁰P₄ which also counts numbers starting with 0.

Spot the difference

PointMultiplication principleAddition principle
Wordsand, thenor, only one
Operationm × nm + n
Examplechoosing a shirt and a trouserone book from 4 maths or 5 science

Practice questions

Easy
  1. 14 shirts and 6 trousers: how many outfits?
  2. 2There are 4 routes by one path or 7 by another: in how many ways can one route be chosen?
  3. 3In how many ways can 3 true-false questions be answered?
Medium
  1. 4How many 2-digit numbers are there in all?
  2. 5How many 3-digit numbers can be formed from 1, 2, 3 with repetition?
  3. 6How many 3-digit numbers are divisible by 5?
Challenging
  1. 7How many 4-digit even numbers have all digits different?
  2. 8How many 3-digit numbers from 1, 2, 3, 4, 5 without repetition have an odd first digit?
Show answers (try them yourself first)
  1. 14 × 6 = 24.
  2. 24 + 7 = 11.
  3. 32 × 2 × 2 = 8.
  4. 49 × 10 = 90.
  5. 53 × 3 × 3 = 27.
  6. 69 × 10 × 2 = 180.
  7. 7एकक 0 असेल तर 9 × 8 × 7 = 504 ; एकक 2, 4, 6, 8 असेल तर 4 × 8 × 8 × 7 = 1792 ; एकूण 2296.
  8. 8पहिला अंक 3 प्रकार (1, 3, 5) ; मग 4 × 3 ; 3 × 12 = 36.

Factorial and permutations

From daily life

When 5 people are lined up there are 5, then 4, 3, 2, 1 choices for the places. Changing the order changes the arrangement. The number of such arrangements is the number of permutations. Building on classes 9–10 : Since classes 9–10 we know products and indices; a factorial is a product of descending numbers.

Simple definition

The factorial is n! = n × (n − 1) × … × 2 × 1 and 0! = 1. The number of ways to take r of n different objects and arrange them in order is the number of permutations ⁿPᵣ.

Rule 1: Permutation formula — ⁿPᵣ = n!/(n − r)! = n(n − 1)…(n − r + 1) (r factors). ⁿPₙ = n!; ⁿP₀ = 1; ⁿP₁ = n.

321= 3! = 6ABCACBBACBCACABCBAⁿPᵣ = n!/(n − r)!

In the figure : Three boxes are filled in 3, 2, 1 ways: 3! = 6 arrangements (ABC, ACB, BAC, BCA, CAB, CBA).

How is the formula obtained? : When r places are filled, the first has n choices, the second n − 1, …, the r-th n − r + 1. The product n(n − 1)…(n − r + 1) becomes n!/(n − r)! on multiplying and dividing by (n − r)!.

  • Easy : 8!/6! = 8 × 7 = 56.
  • Medium : ⁷P₃ = 7 × 6 × 5 = 210.
  • Exam-type : 4-letter arrangements of PLANET: ⁶P₄ = 6 × 5 × 4 × 3 = 360. ⁿP₂ = 42: n(n − 1) = 42, n = 7.
Remember

In permutations order matters: AB and BA are different.

Common mistake

Taking 0! = 0. The correct value is 0! = 1.

Shortcut

In ⁿPᵣ write r factors downward from n: ⁷P₃ = 7 × 6 × 5. Do not expand big factorials fully. Memory clue : "Step down r times from the top".

CET trap

In n! = 5040 the value is n = 7; do not confuse it with 6! = 720 and 8! = 40320.

CET-style example : If ⁿP₂ = 42 then n = ? (A) 7 (B) 6 (C) 8 (D) 21

Answer : (A). n(n − 1) = 42 = 7 × 6. Time-saving method : Two consecutive numbers with product 42: 7 × 6.

Spot the difference

TermMeaningExample
n!arranging all n5! = 120
ⁿPᵣarranging r of n⁶P₄ = 360
ⁿP₀taking nothing1

Practice questions

Easy
  1. 1Find 6!.
  2. 2Find ¹⁰P₂.
  3. 3In how many ways can 5 different books be arranged on a shelf?
Medium
  1. 4Find 9!/(7! × 2!).
  2. 5Find ⁶P₃ + ⁶P₂.
  3. 6If n! = 5040 find n.
Challenging
  1. 7If (n + 1)!/(n − 1)! = 56 find n.
  2. 8If ⁿP₃ = 120 find n.
Show answers (try them yourself first)
  1. 1720.
  2. 210 × 9 = 90.
  3. 35! = 120.
  4. 472/2 = 36.
  5. 5120 + 30 = 150.
  6. 67! = 5040 ; n = 7.
  7. 7(n + 1) × n = 56 = 8 × 7 ; n = 7.
  8. 8n(n − 1)(n − 2) = 120 = 6 × 5 × 4 ; n = 6.

Permutations with repeated objects and restrictions

From daily life

When rearranging the letters of "APPLE", swapping the two P’s does not make a new word, so the arrangements are fewer. There are also conditions like some persons being together or apart. Building on classes 9–10 : Since classes 9–10 we have arranged letters and digits; here we divide for identical objects.

Simple definition

If among n objects p are alike of one kind and q alike of another, the arrangements are n!/(p! × q!). For persons together treat them as one block and multiply by the arrangements inside the block; arrangements with persons apart = total − together.

Rule 1: Alike objects — APPLE: 5!/2! = 60. BANANA: 6!/(3! × 2!) = 60. Divide by the factorial of every repeated letter.

letters A, A, BA1 A2 BA2 A1 B→AABA1 B A2A2 B A1→ABAB A1 A2B A2 A1→BAA3!/2! = 6/2 = 3divide by the factorial of every repeated letter

In the figure : A, A, B: of 6 arrangements two each are the same; the different words are 3 = 3!/2!.

Rule 2: Restrictions — Together: treat as a block and take the factorial of the units × the arrangements inside. A fixed object at a place: fix that place and arrange the rest. When the first digit cannot be 0, fill that place first.

How is the formula obtained? : n!/p!: the p alike objects can be swapped in p! ways without producing a new arrangement, so divide the total n! by p!.

  • Easy : LEVEL: 5!/(2! × 2!) = 120/4 = 30.
  • Medium : DELHI with E and I together: 4! × 2! = 24 × 2 = 48.
  • Exam-type : In 5 people A and B are not together: 5! − 4! × 2! = 120 − 48 = 72. MISSISSIPPI: 11!/(4! × 4! × 2!) = 34650.
Remember

After forming a block do not forget the swaps inside it: 2! or 3!.

Common mistake

Counting the "not together" arrangements directly. It is easier to count the total and the together ones and subtract.

Shortcut

List the repeats first: MISSISSIPPI = I4, S4, P2, M1. Memory clue : "Total factorial divided by the repeats’ factorials".

CET trap

For an even number the last digit must be even, and when 0 is used count the first place separately; fill both conditioned places first.

CET-style example : Arrangements of the letters of TRIANGLE that begin with T and end with E? (A) 720 (B) 5040 (C) 40320 (D) 120

Answer : (A). Both ends fixed, the other 6 letters: 6! = 720. Time-saving method : Once the ends are fixed take the factorial of the rest.

Spot the difference

ConditionMethodExample
Alike lettersn! ÷ factorials of repeatsAPPLE = 60
Togetherblock × insideDELHI, EI together = 48
Not togethertotal − together5 people = 72

Practice questions

Easy
  1. 1Find the arrangements of the letters of INDIA.
  2. 2Find the arrangements of the letters of BOOK.
  3. 3Find the arrangements of MATH that begin with M.
Medium
  1. 4How many 5-digit numbers can be formed from 1, 1, 2, 2, 3?
  2. 5How many 4-digit numbers can be formed from 0, 1, 2, 3, 4 without repetition?
  3. 6How many even 4-digit numbers can be formed from 1, 2, 3, 4, 5 without repetition?
Challenging
  1. 7In how many ways can 3 boys and 3 girls stand alternately?
  2. 8How many arrangements of STUDY have U at neither end?
Show answers (try them yourself first)
  1. 15!/2! = 60.
  2. 24!/2! = 12.
  3. 33! = 6.
  4. 45!/(2! × 2!) = 30.
  5. 5पहिला अंक 4 प्रकार, मग 4 × 3 × 2 ; 4 × 24 = 96.
  6. 6एकक 2 किंवा 4 : 2 ; बाकी 4 × 3 × 2 = 24 ; 48.
  7. 7दोन सुरुवाती × 3! × 3! = 2 × 36 = 72.
  8. 8U ला 3 मधली स्थाने ; बाकी 4! = 24 ; 3 × 24 = 72.

Circular permutations

From daily life

If everyone at a round table moves one seat ahead, the seating pattern stays the same; only who sits next to whom matters. So there are fewer arrangements than in a row. Building on classes 9–10 : Since classes 9–10 we know the circle and its properties; here we count arrangements on a circle.

Simple definition

When n different objects are placed on a circle, arrangements that become equal by a rotation count as one; fixing one object and arranging the rest gives (n − 1)!.

Rule 1: Formulas — Circular arrangements (n − 1)!. A necklace or garland can be turned over, so (n − 1)!/2. If the seats are numbered it is n! as in a row.

ABCDtableA fixedothers: 3! = 6rotation givessame seating(n − 1)!a necklace that can be turned over: (n − 1)!/2

In the figure : Four people round a table: A fixed, the others 3! = 6; if everyone rotates the arrangement is the same.

Rule 2: Restrictions — Two together: treat as a block, (n − 2)! × 2. Not together: (n − 1)! − 2 × (n − 2)!. Alternate: (k − 1)! for one group × k! for the other.

How is the formula obtained? : The n rotations of a circular arrangement are counted as one. The n! row arrangements fall into groups of n that belong to the same circular arrangement; so n!/n = (n − 1)!.

  • Easy : 5 people round a table: (5 − 1)! = 24.
  • Medium : Two together among 7: 2 × 5! = 240.
  • Exam-type : 6 different beads in a necklace: (6 − 1)!/2 = 60. 4 boys and 4 girls alternately: 3! × 4! = 144.
Remember

Row = n!, circle = (n − 1)!, necklace = (n − 1)!/2.

Common mistake

Taking (n − 1)! when the chairs are numbered. There each rotated arrangement is different, so n!.

Shortcut

Fix one person and count as in a row. Memory clue : "Round means one less".

CET trap

(n − 1)!/2 for a necklace holds only when all beads are different; with identical beads count specially.

CET-style example : 8 people round a table with two particular persons not together? (A) 3600 (B) 5040 (C) 1440 (D) 4320

Answer : (A). Total 7! = 5040; together 2 × 6! = 1440; 5040 − 1440 = 3600. Time-saving method : Total minus together.

Spot the difference

TypeFormulaFor 5 objects
Rown!120
Round table(n − 1)!24
Necklace / garland(n − 1)!/212

Practice questions

Easy
  1. 1In how many ways can 3 people sit round a table?
  2. 2In how many ways can 4 different beads be strung into a necklace?
  3. 3In how many ways can 6 people sit on 6 numbered chairs?
Medium
  1. 45 people round a table, two particular persons together: how many ways?
  2. 56 people round a table, two particular persons not together: how many ways?
  3. 6How many garlands can be made of 5 different flowers?
Challenging
  1. 78 people round a table, two particular persons together: how many ways?
  2. 83 couples round a table, every couple together: how many ways?
Show answers (try them yourself first)
  1. 1(3 − 1)! = 2.
  2. 2(4 − 1)!/2 = 3.
  3. 36! = 720.
  4. 42 × 3! = 12.
  5. 55! − 2 × 4! = 120 − 48 = 72.
  6. 6(5 − 1)!/2 = 12.
  7. 72 × 6! = 1440.
  8. 83 गट : (3 − 1)! = 2 ; प्रत्येक जोडपे 2 ; 2 × 2 × 2 × 2 = 16.

Combinations

From daily life

When a committee of 2 is chosen from 4 people, "A and B" and "B and A" are the same; order does not matter. The number of such selections is the number of combinations. Building on classes 9–10 : Since classes 9–10 we know sets and subsets; choosing a subset of r elements is exactly a combination.

Simple definition

The number of ways to choose r of n different objects without regard to order is the number of combinations ⁿCᵣ. If the chosen objects are then to be arranged, multiply by r!.

Rule 1: Formula and properties — ⁿCᵣ = n!/(r! × (n − r)!) = ⁿPᵣ/r!. ⁿCᵣ = ⁿCₙ₋ᵣ; ⁿC₀ = ⁿCₙ = 1; ⁿCᵣ + ⁿCᵣ₊₁ = ⁿ⁺¹Cᵣ₊₁; all subsets 2ⁿ.

ABCD⁴C₂= 6order does not matter:AB = BAⁿCᵣ = n!/(r! (n − r)!) = ⁿPᵣ / r!

In the figure : Choosing 2 of 4 points = 6 lines: ⁴C₂ = 6; AB and BA are the same.

Rule 2: Uses — A particular object included: choose r − 1 from the rest. Excluded: choose r from the rest. "At least one": total − none. "And" means multiply, "or" means add.

How is the formula obtained? : Choosing r objects and arranging them gives ⁿPᵣ; but each selection gives r! arrangements. So ⁿCᵣ = ⁿPᵣ/r! = n!/(r!(n − r)!). Choosing r is the same as leaving out n − r, so ⁿCᵣ = ⁿCₙ₋ᵣ.

  • Easy : ¹⁰C₃ = (10 × 9 × 8)/6 = 120.
  • Medium : ⁸C₅ = ⁸C₃ = (8 × 7 × 6)/6 = 56.
  • Exam-type : 2 of 5 men and 3 of 6 women: ⁵C₂ × ⁶C₃ = 10 × 20 = 200. A committee of 3 from 4 boys and 3 girls with at least one boy: ⁷C₃ − ³C₃ = 35 − 1 = 34.
Remember

When r is large in ⁿCᵣ use ⁿCₙ₋ᵣ: ⁸C₅ = ⁸C₃.

Common mistake

Using permutations to choose a committee. A committee has no order, so use combinations.

Shortcut

Selection = combination, arrangement = permutation; C for a committee, P for a row. Memory clue : "C is for Choose".

CET trap

Counts with and without a particular person differ: for 4 of 9, with him ⁸C₃ = 56, without him ⁸C₄ = 70.

CET-style example : If ⁿC₂ = 28 then n = ? (A) 8 (B) 7 (C) 9 (D) 14

Answer : (A). n(n − 1)/2 = 28, n(n − 1) = 56 = 8 × 7. Time-saving method : Double it and look for two consecutive numbers with that product.

Spot the difference

PointPermutation ⁿPᵣCombination ⁿCᵣ
Ordermattersdoes not matter
Formulan!/(n − r)!n!/(r!(n − r)!)
2 of 63015

Practice questions

Easy
  1. 1Find ¹²C₂.
  2. 2In how many ways can a committee of 3 be chosen from 8 people?
  3. 3If each of 10 people shakes hands once with every other, how many handshakes are there?
Medium
  1. 4If ⁿC₂ = 28 find n.
  2. 5Of 7 points no three are collinear; how many triangles can be formed?
  3. 6A team of 4 from 9 players; how many teams include one particular player?
Challenging
  1. 7How many diagonals does a hexagon have?
  2. 8Find ¹²C₃ + ¹²C₄.
Show answers (try them yourself first)
  1. 112 × 11/2 = 66.
  2. 28 × 7 × 6/6 = 56.
  3. 3¹⁰C₂ = 45.
  4. 4n(n − 1) = 56 ; n = 8.
  5. 5⁷C₃ = 35.
  6. 6⁸C₃ = 56.
  7. 7⁶C₂ − 6 = 15 − 6 = 9.
  8. 8¹²C₃ + ¹²C₄ = ¹³C₄ = 715.

Lesson 3: Permutations and combinations — quick revision

TopicFormula / point
Principlesand = multiply m × n; or = add m + n
Factorialn! = n(n − 1)…1; 0! = 1
PermutationsⁿPᵣ = n!/(n − r)!
Alike objectsn!/(p! q!); together = block × inside; not together = total − together
Circular(n − 1)!; necklace (n − 1)!/2
CombinationsⁿCᵣ = n!/(r!(n − r)!) = ⁿCₙ₋ᵣ
Subsetsall 2ⁿ; ⁿCᵣ + ⁿCᵣ₊₁ = ⁿ⁺¹Cᵣ₊₁

🎯 Most important points for the exam

  • Fill the conditioned place first.
  • P if order matters, C if not.
  • Together means a block and its inner swaps.
  • Round means (n − 1)!.
  • For "at least one" use total − none.
Study mantra

"P for order, C for choice".