Class 11 : Maths (In English) – Lesson 6. Permutations and Combinations
EXPLANATION & SUMMARY
π· Explanation (β1700 words)
π΅ 1οΈβ£ Introduction
In mathematics, Counting Principles help us determine the number of ways in which certain events can occur.
Two key tools for counting are:
π’ Permutations β€ arrangements where order matters
π‘ Combinations β€ selections where order does not matter
π‘ Concept:
When objects are arranged in a specific sequence, we call it a Permutation.
When only selection is important, we call it a Combination.
π΅ 2οΈβ£ Fundamental Principles of Counting
π’ (a) Addition Principle
β‘οΈ If a task A can be done in m ways and task B in n ways, and both cannot occur together,
then total ways = m + n
βοΈ Example:
A student can choose either Science (3 ways) or Commerce (2 ways).
Total ways = 3 + 2 = 5 ways
π’ (b) Multiplication Principle
β‘οΈ If a task A can be done in m ways and B in n ways independently,
then total ways = m Γ n
βοΈ Example:
Selecting 1 shirt (3 types) and 1 pant (4 types)
Total = 3 Γ 4 = 12 combinations
π΅ 3οΈβ£ Permutations (Order Matters)
π’ Definition:
A permutation is an arrangement of r objects from n distinct objects, where order is important.
βοΈ Formula:
P(n, r) = nPα΅£ = n! / (n β r)!
π§ n! (n factorial) = n Γ (nβ1) Γ (nβ2) Γ … Γ 3 Γ 2 Γ 1
βοΈ Example:
Number of 2-digit numbers from digits 1, 2, 3
= 3Pβ = 3! / 1! = 6 arrangements β (12, 13, 21, 23, 31, 32)
π‘ Case 1: All Objects Taken
If r = n β nPβ = n!
π‘ Case 2: Zero Objects
nPβ = 1 (only one arrangement β none selected)
π΅ 4οΈβ£ Permutations with Repetition
When repetition is allowed:
β Each of the r positions can be filled by n objects
Total = nΚ³
βοΈ Example:
3 digits (0, 1, 2) β 2-digit numbers = 3Β² = 9
π΅ 5οΈβ£ Permutations of Distinct Objects
π’ Case (a): All Distinct
Number of permutations = n!
π’ Case (b): Some Objects Identical
If there are n objects with p, q, r,β¦ identical:
β Total permutations = n! / (p! Γ q! Γ r! β¦)
βοΈ Example:
Word βLEVELβ has 5 letters with L repeated 2 times, E repeated 2 times:
Permutations = 5! / (2! Γ 2!) = 30
π΅ 6οΈβ£ Circular Permutations
In a circle, arrangements differing only by rotation are same.
β Formula (distinct objects):
(n β 1)!
β Formula (if clockwise β anticlockwise):
n! / 2
βοΈ Example:
Arranging 4 persons around a round table = (4β1)! = 6 ways
π΅ 7οΈβ£ Combinations (Order Doesnβt Matter)
π’ Definition:
A combination is a selection of r objects from n objects where order is irrelevant.
βοΈ Formula:
C(n, r) = nCα΅£ = n! / [r! Γ (n β r)!]
βοΈ Example:
Choose 2 fruits from {apple, banana, mango}
nCα΅£ = 3Cβ = 3! / (2! Γ 1!) = 3 combinations
π§ Set: {A, B, M} β {A,B}, {A,M}, {B,M}
π‘ Relation Between Permutations and Combinations
nPα΅£ = nCα΅£ Γ r!
βοΈ Example:
5Pβ = 5Cβ Γ 3! = 10 Γ 6 = 60 β
π΅ 8οΈβ£ Properties of Combinations
1οΈβ£ nCβ = 1
2οΈβ£ nCβ = 1
3οΈβ£ nCα΅£ = nCββα΅£
4οΈβ£ nCα΅£ + nCα΅£ββ = (n + 1)Cα΅£
π΅ 9οΈβ£ Applications
π’ (a) Selection and Arrangement
Choose 3 persons from 5 for a committee:
Ways = 5Cβ = 10
Arrange 3 persons in order:
Ways = 5Pβ = 60
π’ (b) Forming Numbers from Digits
Using 1, 2, 3 β 2-digit numbers = 3Pβ = 6
π’ (c) Forming Teams
Select 11 players from 15 β 15Cββ
π’ (d) Probability Calculations
Permutations/Combinations often used in probability to count outcomes.
π΅ π Factorial Function
Used heavily in formulas.
βοΈ n! = n Γ (nβ1)!
βοΈ 0! = 1
βοΈ Example:
5! = 120
4! = 24
π΅ π΅ 11. Difference Between Permutation and Combination (Plain Text)
πΉ Permutation:
Order or sequence of arrangement is important.
Formula: nPα΅£ = n! / (n β r)!
Example: If we arrange A and B, then AB and BA are different permutations.
πΉ Combination:
Order or sequence is not important.
Formula: nCα΅£ = n! / (r! Γ (n β r)!)
Example: If we choose A and B, then AB and BA are same combination.
π‘ Key Point:
Permutation = Arrangement (Order matters)
Combination = Selection (Order doesnβt matter)
βοΈ Example:
If we have 3 letters A, B, C and we choose 2 at a time:
β’ Permutations: AB, BA, AC, CA, BC, CB β total 6
β’ Combinations: {A,B}, {A,C}, {B,C} β total 3
β Conclusion:
π Use Permutation when the arrangement or order matters.
π Use Combination when only selection matters.
π΅ 1οΈβ£2οΈβ£ Mixed Cases
π’ Example 1:
Number of ways to form 3-letter words from A, B, C, D
(Without repetition): 4Pβ = 24
(With repetition): 4Β³ = 64
π’ Example 2:
Form 3-digit even numbers using {1,2,3,4,5,6}
Units place even = 3 choices (2,4,6)
Hundreds place = 5, Tens = 4
Total = 3 Γ 5 Γ 4 = 60
π΅ 1οΈβ£3οΈβ£ Solving Step-by-Step
π’ Example:
How many 3-letter words (meaningful or not) from letters A,B,C,D if repetition not allowed?
β‘ Step 1: n = 4, r = 3
β‘ Step 2: nPα΅£ = 4! / (4β3)! = 24
β Answer: 24
π΅ 1οΈβ£4οΈβ£ Special Identities
β nCβ = n
β nCβ = n(nβ1)/2
β nCβ = n(nβ1)(nβ2)/6
βοΈ Example:
7Cβ = 7Γ6/2 = 21
π΅ 1οΈβ£5οΈβ£ Use in Real Life
π Arrangements: Seating, passwords, codes
π Selections: Teams, committees
π Probability: Counting favorable outcomes
π΅ 1οΈβ£6οΈβ£ Important Tips
β Always identify if order matters β Permutation
β If only selection matters β Combination
β Check for repetition or restriction
β Apply factorial simplification carefully
π· Summary (β300 words)
πΉ Counting Principle
If events occur in sequence: Multiply
If events occur exclusively: Add
πΉ Permutation
Arrangement of r objects from n
Formula: nPα΅£ = n! / (n β r)!
If all taken: nPβ = n!
If none: nPβ = 1
πΉ With repetition: nΚ³
πΉ Circular permutations:
(nβ1)! if rotations same; n!/2 if rotations distinct
πΉ Combination
Selection where order not important
Formula: nCα΅£ = n! / (r!(n β r)!)
Relation: nPα΅£ = nCα΅£ Γ r!
πΉ Properties:
nCβ = 1, nCβ = 1, nCα΅£ = nCββα΅£,
nCα΅£ + nCα΅£ββ = (n+1)Cα΅£
πΉ Examples:
5Cβ = 10, 5Pβ = 60
πΉ Factorial Rules:
n! = n Γ (nβ1)!; 0! = 1
πΉ Identities:
nCβ = n, nCβ = n(nβ1)/2
πΉ When to Use:
π§ Permutation β order, arrangement
π§ Combination β selection, group formation
πΉ Applications:
β’ Number of ways to arrange letters
β’ Select committees
β’ Probability problems
π Quick Recap
β Order matters β Permutation
β Order doesnβt matter β Combination
β Repetition allowed β nΚ³
β Factorial basics β 0! = 1
β nPα΅£ = nCα΅£ Γ r!
β Circle arrangements β (nβ1)!
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QUESTIONS FROM TEXTBOOK
Exercise 6.1
π΅ Question 1
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?
π’ Answer
β‘οΈ (i) Repetition allowed: choices for hundreds = 5, tens = 5, units = 5
β‘οΈ Total = 5 Γ 5 Γ 5 = 5Β³ = 125
βοΈ Answer (i): 125
β‘οΈ (ii) Repetition not allowed: number of permutations = 5P3
β‘οΈ 5P3 = 5 Γ 4 Γ 3 = 60
βοΈ Answer (ii): 60
π΅ Question 2
How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?
π’ Answer
β‘οΈ Even number β units place must be one of {2, 4, 6} β 3 choices
β‘οΈ Hundreds place: any of 6 digits (repetition allowed) β 6 choices
β‘οΈ Tens place: any of 6 digits β 6 choices
β‘οΈ Total = 6 Γ 6 Γ 3 = 108
βοΈ Answer: 108
π΅ Question 3
How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?
π’ Answer
β‘οΈ βFirst 10 lettersβ = 10 distinct symbols
β‘οΈ No repetition β permutations 10P4
β‘οΈ 10P4 = 10 Γ 9 Γ 8 Γ 7 = 5040
βοΈ Answer: 5040
π΅ Question 4
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?
π’ Answer
β‘οΈ First two places fixed as β6β and β7β (already used)
β‘οΈ Remaining digits available = {0,1,2,3,4,5,8,9} β 8 digits
β‘οΈ Remaining 3 places filled without repetition β 8P3
β‘οΈ 8P3 = 8 Γ 7 Γ 6 = 336
βοΈ Answer: 336
π΅ Question 5
A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?
π’ Answer
β‘οΈ Each toss has 2 outcomes (H/T)
β‘οΈ Independent tosses β total outcomes = 2Β³ = 8
βοΈ Answer: 8
π΅ Question 6
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?
π’ Answer
β‘οΈ Order matters (topβbottom different) β permutations of 5 taken 2
β‘οΈ 5P2 = 5 Γ 4 = 20
βοΈ Answer: 20
π Exercise 6.2
π΅ Question 1:
Evaluate
(i) 8!
(ii) 4! β 3!
π’ Answer:
(i)
β‘οΈ 8! = 8 Γ 7 Γ 6 Γ 5 Γ 4 Γ 3 Γ 2 Γ 1
β‘οΈ 8! = 40320
βοΈ Answer: 40320
(ii)
β‘οΈ 4! β 3! = (4 Γ 3 Γ 2 Γ 1) β (3 Γ 2 Γ 1)
β‘οΈ 24 β 6 = 18
βοΈ Answer: 18
π΅ Question 2:
Is 3! + 4! = 7!?
π’ Answer:
β‘οΈ 3! = 3 Γ 2 Γ 1 = 6
β‘οΈ 4! = 4 Γ 3 Γ 2 Γ 1 = 24
β‘οΈ 3! + 4! = 6 + 24 = 30
β‘οΈ 7! = 7 Γ 6 Γ 5 Γ 4 Γ 3 Γ 2 Γ 1 = 5040
βοΈ 30 β 5040
βοΈ No, 3! + 4! β 7!
π΅ Question 3:
Compute 8! / (6! Γ 2!)
π’ Answer:
β‘οΈ 8! / (6! Γ 2!)
β‘οΈ = (8 Γ 7 Γ 6!) / (6! Γ 2 Γ 1)
β‘οΈ = (8 Γ 7) / 2
β‘οΈ = 56 / 2 = 28
βοΈ Answer: 28
π΅ Question 4:
If 1/6! + 1/7! = x / 8!, find x
π’ Answer:
β‘οΈ Take LCM 7!
β‘οΈ 1/6! + 1/7! = (7 + 1) / 7! = 8 / 7!
β‘οΈ We know 8! = 8 Γ 7!
β‘οΈ So 8 / 7! = (8 Γ 8) / 8! = 64 / 8!
βοΈ x = 64
π΅ Question 5:
Evaluate n! / (n β r)!, when
(i) n = 6, r = 2
(ii) n = 9, r = 5
π’ Answer:
(i)
β‘οΈ 6! / (6 β 2)! = 6! / 4!
β‘οΈ = (6 Γ 5 Γ 4!) / 4! = 6 Γ 5 = 30
βοΈ Answer: 30
(ii)
β‘οΈ 9! / (9 β 5)! = 9! / 4!
β‘οΈ = (9 Γ 8 Γ 7 Γ 6 Γ 5 Γ 4!) / 4!
β‘οΈ = 9 Γ 8 Γ 7 Γ 6 Γ 5
β‘οΈ = 15120
βοΈ Answer: 15120
π Exercise 6.3
π΅ Question 1
How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
π’ Answer
β‘οΈ Hundreds place: 9 choices (1β9)
β‘οΈ Tens place: 8 choices (remaining)
β‘οΈ Units place: 7 choices (remaining)
β‘οΈ Total = 9 Γ 8 Γ 7 = 504
βοΈ Final: 504
π΅ Question 2
How many 4-digit numbers are there with no digit repeated?
π’ Answer
β‘οΈ Digits available: 0β9 (10 digits)
β‘οΈ Thousands place: 9 choices (1β9; 0 not allowed)
β‘οΈ Hundreds place: 9 choices (0 plus remaining 8)
β‘οΈ Tens place: 8 choices
β‘οΈ Units place: 7 choices
β‘οΈ Total = 9 Γ 9 Γ 8 Γ 7 = 4536
βοΈ Final: 4536
π΅ Question 3
How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?
π’ Answer
β‘οΈ Units (even): {2, 4, 6} β 3 choices
β‘οΈ Hundreds: from remaining 5 digits β 5 choices
β‘οΈ Tens: remaining β 4 choices
β‘οΈ Total = 3 Γ 5 Γ 4 = 60
βοΈ Final: 60
π΅ Question 4
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?
π’ Answer (all 4-digit, no repetition)
β‘οΈ Total = 5P4 = 5 Γ 4 Γ 3 Γ 2 = 120
π’ Answer (even)
β‘οΈ Units must be even: {2, 4} β 2 choices
β‘οΈ Thousands: from remaining 4 digits β 4 choices
β‘οΈ Hundreds: 3 choices
β‘οΈ Tens: 2 choices
β‘οΈ Even count = 2 Γ 4 Γ 3 Γ 2 = 48
βοΈ Final: Total 120; Even 48
π΅ Question 5
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?
π’ Answer
β‘οΈ Ordered positions (distinct posts) β permutations
β‘οΈ 8P2 = 8 Γ 7 = 56
βοΈ Final: 56
π΅ Question 6
Find n if (nβ1)P3 : nP4 = 1 : 9.
π’ Answer
β‘οΈ (nβ1)P3 = (nβ1)!/(nβ4)!
β‘οΈ nP4 = n!/(nβ4)!
β‘οΈ Ratio = [(nβ1)!/(nβ4)!] Γ· [n!/(nβ4)!] = (nβ1)!/n! = 1/n
β‘οΈ 1/n = 1/9 β n = 9
βοΈ Final: n = 9
π΅ Question 7
Find r if
(i) β΅Pα΅£ = 2 Γ βΆPα΅£ββ
(ii) β΅Pα΅£ = βΆPα΅£ββ
π’ Answer (i):
β‘οΈ Formula: βΏPα΅£ = n! / (n β r)!
So,
5! / (5 β r)! = 2 Γ 6! / (6 β r + 1)!
β 120 / (5 β r)! = 2 Γ 720 / (7 β r)!
β‘οΈ Multiply both sides by (5 β r)!:
120 Γ (7 β r)(6 β r) = 2 Γ 720
β‘οΈ Simplify: (7 β r)(6 β r) = 12
Expand: rΒ² β 13r + 42 = 12
β‘οΈ rΒ² β 13r + 30 = 0
Factor: (r β 10)(r β 3) = 0
βοΈ So, r = 3 or 10 (not possible > 5)
π’ Final: r = 3
π’ Answer (ii):
5! / (5 β r)! = 6! / (6 β r + 1)!
β 120 / (5 β r)! = 720 / (7 β r)!
Multiply by (5 β r)!:
120 Γ (7 β r)(6 β r) = 720
Simplify: (7 β r)(6 β r) = 6
Expand: rΒ² β 13r + 42 = 6
β‘οΈ rΒ² β 13r + 36 = 0
Factor: (r β 9)(r β 4) = 0
βοΈ r = 4 or 9 (r β€ 5 β r = 4)
π’ Final: r = 4
π΅ Question 8
How many words (with or without meaning) can be formed using all letters of the word EQUATION, using each letter exactly once?
π’ Answer:
Total letters = 8 (all distinct)
β‘οΈ Number of arrangements = 8! = 40320
βοΈ Final Answer: 40320 words
π΅ Question 9
How many words (with or without meaning) can be formed from the letters of the word MONDAY, assuming no letter repeated:
(i) 4 letters at a time
(ii) All letters used
(iii) All letters used but first letter is a vowel
π’ Answer (i):
Total letters = 6
β‘οΈ Number of 4-letter arrangements = βΆPβ = 6 Γ 5 Γ 4 Γ 3 = 360
π’ Answer (ii):
All letters used β βΆPβ = 6! = 720
π’ Answer (iii):
Vowels = O, A β 2 choices for 1st position
Remaining 5 letters can be arranged in 5! = 120 ways
Total = 2 Γ 120 = 240
βοΈ Final Answers: (i) 360, (ii) 720, (iii) 240
π΅ Question 10
In how many of the distinct permutations of letters in MISSISSIPPI do the four Iβs not come together?
π’ Answer:
Total letters = 11 (M-1, I-4, S-4, P-2)
β‘οΈ Total permutations = 11! / (4!Γ4!Γ2!) = 34650
β‘οΈ When 4 Iβs together β treat as one unit
Then letters = [IIII], M, S, S, S, S, P, P β 8 total
Arrangements = 8! / (4!Γ2!) = 840
β‘οΈ Required = 34650 β 840 = 33810
βοΈ Final Answer: 33810
π΅ Question 11
How many ways can letters of PERMUTATIONS be arranged if
(i) Words start with P and end with S
(ii) Vowels are all together
(iii) There are always 4 letters between P and S
π’ Answer (i):
Fix P at start, S at end
Remaining 10 letters arranged = 10! = 3628800
π’ Answer (ii):
Vowels = E, U, A, I, O (5 vowels) β treat as one unit
Now total = 7 units (vowel block + 7 consonants)
Arrangements = 7! Γ 5! = 5040 Γ 120 = 604800
π’ Answer (iii):
Distance between P and S = 4 letters
Position choices:
P can be at 1 to 7
For each, 2 orders (P before S or S before P)
β‘οΈ Total = 7 Γ 2 = 14 ways to place P, S
Remaining 9 letters = 9! ways
Total = 14 Γ 9! = 14 Γ 362880 = 5080320
βοΈ Final Answers:
(i) 3628800
(ii) 604800
(iii) 5080320
π§ Exercise 6.4
π΅ Question 1:
If βΏCβ = βΏCβ, find βΏCβ.
π’ Answer:
πΉ Property: βΏCα΅£ = βΏCββα΅£
πΉ Given: βΏCβ = βΏCβ β 8 = n β 2
πΉ Solve:
β8 = n β 2
βn = 10
πΉ Compute:
βΒΉβ°Cβ = (10 Γ 9)/2 = 45
βοΈ Final Answer: βΏCβ = 45
π΅ Question 2:
Determine n if
(i) Β²βΏCβ : βΏCβ = 12 : 1
(ii) Β²βΏCβ : βΏCβ = 11 : 1
π’ (i) Answer:
πΉ Formula:
βΒ²βΏCβ / βΏCβ = [2n(2n β 1)(2n β 2)] / [n(n β 1)(n β 2)]
πΉ Simplify:
β= 4(2n β 1)/(n β 2)
πΉ Equation:
β4(2n β 1)/(n β 2) = 12
πΉ Solve:
β2n β 1 = 3(n β 2)
β2n β 1 = 3n β 6
βn = 5
βοΈ Final: n = 5
π’ (ii) Answer:
πΉ Equation: 4(2n β 1)/(n β 2) = 11
πΉ Solve:
β8n β 4 = 11n β 22
β3n = 18
βn = 6
βοΈ Final: n = 6
π΅ Question 3:
How many chords can be drawn through 21 points on a circle?
π’ Answer:
πΉ Each chord determined by any 2 points
πΉ Required number = Β²ΒΉCβ = (21 Γ 20)/2 = 210
βοΈ Final Answer: 210
π΅ Question 4:
In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
π’ Answer:
πΉ Select boys: β΅Cβ = 10
πΉ Select girls: β΄Cβ = 4
πΉ Total = 10 Γ 4 = 40
βοΈ Final Answer: 40 ways
π΅ Question 5:
Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.
π’ Answer:
πΉ Red: βΆCβ = 20
πΉ White: β΅Cβ = 10
πΉ Blue: β΅Cβ = 10
πΉ Multiply: 20 Γ 10 Γ 10 = 2000
βοΈ Final Answer: 2000
π΅ Question 6:
Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
π’ Answer:
πΉ Choose 1 Ace: β΄Cβ = 4
πΉ Choose 4 non-Aces: β΄βΈCβ
β= (48 Γ 47 Γ 46 Γ 45)/(4 Γ 3 Γ 2 Γ 1)
β= 194580
πΉ Multiply: 4 Γ 194580 = 778320
βοΈ Final Answer: 778320
π΅ Question 7:
In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?
π’ Answer:
πΉ Bowlers: β΅Cβ = 5
πΉ Others: ΒΉΒ²Cβ = ΒΉΒ²Cβ
= 792
πΉ Total = 5 Γ 792 = 3960
βοΈ Final Answer: 3960
π΅ Question 8:
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.
π’ Answer:
πΉ Black: β΅Cβ = 10
πΉ Red: βΆCβ = 20
πΉ Total = 10 Γ 20 = 200
βοΈ Final Answer: 200
π΅ Question 9:
In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
π’ Answer:
πΉ Remaining courses to choose = 5 β 2 = 3
πΉ Remaining available = 9 β 2 = 7
πΉ Ways = β·Cβ = 35
βοΈ Final Answer: 35
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OTHER IMPORTANT QUESTIONS FOR EXAMS
CBSE STYLE BOARD PAPER
ESPECIALLY FROM THIS CHAPTER ONLY
π· Section A β Multiple Choice Questions (1 mark each)
π΅ Question 1:
The value of 5Pβ is
π’ (A) 10
π΅ (B) 15
π (C) 20
π΄ (D) 30
βοΈ Answer: (D) 30
π‘ 5Pβ = 5! / (5β2)! = (5Γ4Γ3!)/3! = 20 β
Correction: Answer = (C) 20
π΅ Question 2:
The value of 6Cβ is
π’ (A) 10
π΅ (B) 15
π (C) 20
π΄ (D) 30
βοΈ Answer: (B) 15
π‘ 6Cβ = 6! / (2!Γ4!) = (6Γ5)/2 = 15
π΅ Question 3:
Which of the following is true?
π’ (A) nPβ = 0
π΅ (B) nPβ = 1
π (C) nPβ = n
π΄ (D) nPβ = n!
βοΈ Answer: (B) nPβ = 1
π΅ Question 4:
If nPβ = 56, find n.
π’ (A) 7
π΅ (B) 8
π (C) 9
π΄ (D) 6
βοΈ Answer: (B) 8
π‘ nPβ = n(nβ1) = 56 β nΒ² β n β 56 = 0 β n = 8
π΅ Question 5:
nCβ = 10, find n.
π’ (A) 5
π΅ (B) 6
π (C) 4
π΄ (D) 7
βοΈ Answer: (B) 5? Check:
nCβ = n(nβ1)/2 = 10 β nΒ² β n β 20 = 0 β n = 5 β
Answer = (A) 5
π΅ Question 6:
The number of ways to arrange 4 distinct objects =
π’ (A) 4
π΅ (B) 4Β²
π (C) 4!
π΄ (D) 2β΄
βοΈ Answer: (C) 4! = 24
π΅ Question 7:
The number of ways to select 3 students from 6 =
π’ (A) 6Cβ
π΅ (B) 6Pβ
π (C) 6!
π΄ (D) 3!
βοΈ Answer: (A) 6Cβ = 20
π΅ Question 8:
nCα΅£ = nCββα΅£ means
π’ (A) Selection of r or (nβr) are same
π΅ (B) Selection of r or (nβr) different
π (C) Always greater
π΄ (D) None
βοΈ Answer: (A) Selections are same
π΅ Question 9:
If 0! = ?
π’ (A) 0
π΅ (B) 1
π (C) Undefined
π΄ (D) n
βοΈ Answer: (B) 1
π΅ Question 10:
The number of 3-digit numbers using digits 1,2,3,4 (no repetition)
π’ (A) 4Pβ
π΅ (B) 4Cβ
π (C) 4Β³
π΄ (D) 3!
βοΈ Answer: (A) 4Pβ = 24
π΅ Question 11:
If order matters, use
π’ (A) Combination
π΅ (B) Permutation
π (C) Both
π΄ (D) None
βοΈ Answer: (B) Permutation
π΅ Question 12:
If order doesnβt matter, use
π’ (A) Permutation
π΅ (B) Combination
π (C) Both
π΄ (D) None
βοΈ Answer: (B) Combination
π΅ Question 13:
The value of 5Cβ is
π’ (A) 10
π΅ (B) 20
π (C) 15
π΄ (D) 25
βοΈ Answer: (A) 10
π΅ Question 14:
Relation between nPα΅£ and nCα΅£ is
π’ (A) nPα΅£ = nCα΅£ Γ r!
π΅ (B) nPα΅£ = nCα΅£ / r!
π (C) nPα΅£ = nCα΅£ + r!
π΄ (D) nPα΅£ = nCα΅£
βοΈ Answer: (A) nPα΅£ = nCα΅£ Γ r!
π΅ Question 15:
The number of permutations of letters in βLEVELβ =
π’ (A) 5!
π΅ (B) 5! / (2!2!)
π (C) 5! / 2!
π΄ (D) 5!/3!
βοΈ Answer: (B) 5! / (2!2!) = 30
π΅ Question 16:
The number of circular permutations of 5 distinct items =
π’ (A) 5!
π΅ (B) 4!
π (C) 3!
π΄ (D) 2!
βοΈ Answer: (B) 4! = 24
π΅ Question 17:
In how many ways can 3 letters be arranged from A,B,C,D?
π’ (A) 4Pβ
π΅ (B) 4Cβ
π (C) 3!
π΄ (D) 4Β³
βοΈ Answer: (A) 4Pβ = 24
π΅ Question 18:
The value of 8Pβ is
π’ (A) 0
π΅ (B) 1
π (C) 8
π΄ (D) 8!
βοΈ Answer: (B) 1
π· Section B β Short Answer Type (2β3 marks each)
π΅ Question 19:
Find the value of 7Pβ.
π’ Answer:
β‘ Step 1: Formula β nPα΅£ = n! / (n β r)!
β‘ Step 2: Substitute β 7Pβ = 7! / 4!
β‘ Step 3: Simplify β (7 Γ 6 Γ 5 Γ 4!) / 4! = 7 Γ 6 Γ 5 = 210
β Final Answer: 210
π΅ Question 20:
Find the value of 10Cβ.
π’ Answer:
β‘ Step 1: Formula β nCα΅£ = n! / [r!(n β r)!]
β‘ Step 2: Substitute β 10Cβ = 10! / (4! Γ 6!)
β‘ Step 3: Simplify β (10 Γ 9 Γ 8 Γ 7) / (4 Γ 3 Γ 2 Γ 1) = 5040 / 24 = 210
β Final Answer: 210
π΅ Question 21:
How many 3-digit numbers can be formed using digits 1, 2, 3, 4, 5 (without repetition)?
π’ Answer:
β‘ Step 1: n = 5 digits, r = 3
β‘ Step 2: Formula β nPα΅£ = n! / (n β r)!
β‘ Step 3: 5Pβ = 5! / 2! = (5 Γ 4 Γ 3) = 60
β Final Answer: 60 numbers
π΅ Question 22:
Find the number of ways to select a team of 3 students from 8 students.
π’ Answer:
β‘ Step 1: Formula β nCα΅£ = n! / [r!(n β r)!]
β‘ Step 2: Substitute β 8Cβ = 8! / (3! Γ 5!)
β‘ Step 3: Simplify β (8 Γ 7 Γ 6) / (3 Γ 2 Γ 1) = 336 / 6 = 56
β Final Answer: 56 ways
π΅ Question 23:
Find the number of ways to arrange letters of the word βAPPLEβ.
π’ Answer:
β‘ Step 1: Total letters = 5 (A, P, P, L, E)
β‘ Step 2: Repetition β P repeated 2 times
β‘ Step 3: Formula β n! / (p!) = 5! / 2! = (120 / 2) = 60
β Final Answer: 60 arrangements
π· Section C β Mid-Length Questions (3 Marks Each)
π΅ Question 24:
In how many ways can the letters of the word βLEVELβ be arranged?
π’ Answer:
β‘ Step 1: Letters = 5 (L, E, V, E, L)
β‘ Step 2: Repetitions β L = 2, E = 2
β‘ Step 3: Formula β n! / (p! Γ q!)
β‘ Step 4: 5! / (2! Γ 2!) = 120 / 4 = 30
β Final Answer: 30 arrangements
π΅ Question 25:
Find the number of 4-letter words formed using letters A, B, C, D, E if repetition is allowed.
π’ Answer:
β‘ Step 1: n = 5 letters, r = 4
β‘ Step 2: Repetition allowed β total = nΚ³
β‘ Step 3: 5β΄ = 625
β Final Answer: 625 words
π΅ Question 26:
Find number of permutations of 5 objects taken all at a time.
π’ Answer:
β‘ Step 1: Formula β nPβ = n!
β‘ Step 2: 5Pβ
= 5! = 120
β Final Answer: 120
π΅ Question 27:
Find number of combinations of 10 objects taken all at a time.
π’ Answer:
β‘ Step 1: Formula β nCβ = 1
β‘ Step 2: 10Cββ = 1
β Final Answer: 1
π· Section D β Long Answer Type (5 Marks Each)
π΅ Question 28:
Find the number of permutations of the letters of the word βMISSISSIPPIβ.
π’ Answer:
β‘ Step 1: Total letters = 11
β‘ Step 2: Repetitions β M = 1, I = 4, S = 4, P = 2
β‘ Step 3: Formula β n! / (p! Γ q! Γ r! Γ s!)
β‘ Step 4: 11! / (1! Γ 4! Γ 4! Γ 2!)
β‘ Step 5: Simplify β 39916800 / (1 Γ 24 Γ 24 Γ 2)
β‘ Step 6: 39916800 / 1152 = 34650
β Final Answer: 34650 distinct arrangements
π΅ Question 29:
How many 4-digit numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if repetition is not allowed?
π’ Answer:
β‘ Step 1: n = 6 digits, r = 4
β‘ Step 2: Formula β nPα΅£ = n! / (n β r)!
β‘ Step 3: 6Pβ = 6! / 2! = (6 Γ 5 Γ 4 Γ 3) = 360
β Final Answer: 360 different 4-digit numbers
π΅ Question 30:
In how many ways can a committee of 4 persons be formed from 7 men and 5 women, if the committee includes at least 2 women?
π’ Answer:
β‘ Step 1: Total = 7 men + 5 women
β‘ Step 2: We need 4 members with β₯ 2 women
πΈ Case 1: 2 women + 2 men β 5Cβ Γ 7Cβ = 10 Γ 21 = 210
πΈ Case 2: 3 women + 1 man β 5Cβ Γ 7Cβ = 10 Γ 7 = 70
πΈ Case 3: 4 women + 0 men β 5Cβ Γ 7Cβ = 5 Γ 1 = 5
β‘ Step 3: Total = 210 + 70 + 5 = 285
β Final Answer: 285 ways
π΅ Question 31:
A question paper has 10 questions. In how many ways can a student select 6 questions, if he must answer at least 2 from first 5?
π’ Answer:
β‘ Step 1: Divide into two groups
Group A = first 5 questions, Group B = last 5 questions
β‘ Step 2: Cases
πΈ Case 1: 2 from A and 4 from B β 5Cβ Γ 5Cβ = 10 Γ 5 = 50
πΈ Case 2: 3 from A and 3 from B β 5Cβ Γ 5Cβ = 10 Γ 10 = 100
πΈ Case 3: 4 from A and 2 from B β 5Cβ Γ 5Cβ = 5 Γ 10 = 50
πΈ Case 4: 5 from A and 1 from B β 5Cβ
Γ 5Cβ = 1 Γ 5 = 5
β‘ Step 3: Total = 50 + 100 + 50 + 5 = 205
β Final Answer: 205 ways
π· Section E β Case Study / Application Based (5 Marks Each)
π΅ Question 32 (Case Study):
A school has to form a team of 3 students for a quiz from 5 boys and 4 girls.
Find number of ways if:
(a) At least 1 girl is selected
(b) All boys are selected
π’ Answer:
β‘ Total students = 9
β‘ (a) At least 1 girl
πΈ Total combinations = 9Cβ = 84
πΈ All boys = 5Cβ = 10
β‘ So combinations with at least 1 girl = 84 β 10 = 74
β Answer (a): 74 ways
β‘ (b) All boys
πΈ 5Cβ = 10
β Answer (b): 10 ways
π΅ Question 33 (Application):
A lock consists of 3 rings, each containing digits 0 to 9.
(a) How many possible lock codes are there if repetition is allowed?
(b) How many codes if no digit is repeated?
π’ Answer:
β‘ (a) Repetition allowed β each ring = 10 choices
Total = 10 Γ 10 Γ 10 = 1000 codes
β‘ (b) No repetition β 10 Γ 9 Γ 8 = 720 codes
β Final Answers:
(a) 1000 codes
(b) 720 codes
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