CBSE Class 11 Maths Revision Notes Chapter 6 Permutations and Combinations
Permutations and Combinations explains how to count arrangements and selections without listing every possibility. For CBSE Class 11 Maths 2026–27, this chapter covers counting principles, factorial notation, permutations, combinations and formula-based questions.
Think of arranging students in a row, forming numbers from given digits or selecting a team from a group. In some questions, the order matters. In others, only selection matters. This chapter teaches how to identify both cases and use the correct formula.
Use these CBSE Class 11 Maths Revision Notes Chapter 6 to revise the fundamental principle of counting, multiplication principle, addition principle, factorial notation, permutation formula, combination formula and common application-based questions.
These Class 11 Maths Chapter 6 Notes are useful for quick revision before solving practice questions from Permutations and Combinations.
Key Takeaways
- Counting principles: Addition and multiplication principles help count outcomes step by step.
- Permutation: A permutation is an arrangement where order matters.
- Combination: A combination is a selection where order does not matter.
- Formulas: nPr = n!/(n - r)! and nCr = n!/[r!(n - r)!] are the main formulas.
Need help with Permutations and Combinations?
Access interactive practice, live doubt-solving sessions and complete NCERT-based support on the Extramarks Learning App. Sign Up Free
Class 11 Maths Chapter 6 Notes for Permutations and Combinations Revision
These notes are arranged for 30-minute revision, so students can revise the chapter in a simple order.
Start with the fundamental principle of counting. Then revise factorial notation. After that, understand permutations and combinations separately. Finally, practise how to decide whether a question is based on arrangement or selection.
| Topic | What You Revise |
| Fundamental principle of counting | Counting outcomes without listing them |
| Addition principle | Cases where one option or another can happen |
| Multiplication principle | Step-by-step choices happening together |
| Factorial notation | Meaning and use of n! |
| Permutation | Arrangement where order matters |
| Combination | Selection where order does not matter |
| nPr formula | Formula for arrangements |
| nCr formula | Formula for selections |
| Word arrangements | Counting arrangements of letters |
| Number formation | Forming numbers using given digits |
| Committee selection | Selecting people from a group |
What is Counting in Mathematics?
Counting in this chapter does not mean counting objects one by one.
It means finding the number of possible ways in which an event can happen.
Example:
How many 3-digit numbers can be formed using the digits 1, 2, 3 and 4 without repetition?
Instead of listing all numbers, we count the choices for each place.
Hundreds place = 4 choices
Tens place = 3 choices
Units place = 2 choices
Total numbers = 4 × 3 × 2 = 24
This is the basic idea behind permutations and combinations.
Fundamental Principle of Counting
The fundamental principle of counting helps find the total number of ways when a task has different steps or cases.
It has two main rules:
| Principle | Used When |
| Addition principle | One of many separate cases can happen |
| Multiplication principle | A task happens in steps |
These two rules are the base of Class 11 Maths Permutations and Combinations Notes.
Addition Principle
The addition principle is used when one task can be done in one way or another way, but not both at the same time.
If one task can be done in m ways and another task can be done in n ways, then either task can be done in:
m + n ways
Example:
A student can travel to school by bus in 3 ways or by auto in 2 ways.
Total choices = 3 + 2 = 5
Use addition when the cases are separate.
Multiplication Principle
The multiplication principle is used when a task is completed in steps.
If the first step can be done in m ways and the second step can be done in n ways, then both steps together can be done in:
m × n ways
Example:
A student has 3 shirts and 2 trousers.
Number of outfits = 3 × 2 = 6
Use multiplication when each choice is followed by another choice.
Addition Principle vs Multiplication Principle
| Basis | Addition Principle | Multiplication Principle |
| Meaning | One case or another case | One step followed by another step |
| Keyword | Or | And |
| Formula | m + n | m × n |
| Example | Choose tea or coffee | Choose shirt and trousers |
Many counting errors happen when students use addition instead of multiplication, or the other way around.
Factorial Notation
Factorial notation is used to write products of consecutive positive integers.
For a positive integer n:
n! = n × (n - 1) × (n - 2) × ... × 3 × 2 × 1
Examples:
5! = 5 × 4 × 3 × 2 × 1 = 120
4! = 4 × 3 × 2 × 1 = 24
3! = 3 × 2 × 1 = 6
Also:
0! = 1
1! = 1
Factorial notation is used in both nPr formula and nCr formula.
Important Factorial Values
| n | n! |
| 0 | 1 |
| 1 | 1 |
| 2 | 2 |
| 3 | 6 |
| 4 | 24 |
| 5 | 120 |
| 6 | 720 |
| 7 | 5040 |
Students should revise these values before solving formula-based questions.
What is a Permutation?
A permutation is an arrangement of objects in a particular order.
Order matters in permutations.
Example:
Suppose A, B and C are three students.
Arrangements of two students are:
AB, BA, AC, CA, BC, CB
Here, AB and BA are different because the order is different.
So, this is a permutation problem.
Permutation Formula
The number of permutations of n different objects taken r at a time is:
nPr = n!/(n - r)!
Here:
n = total number of objects
r = number of objects arranged
Condition:
0 ≤ r ≤ n
Example:
Find the number of ways of arranging 3 students from 5 students.
n = 5
r = 3
5P3 = 5!/(5 - 3)!
= 5!/2!
= 5 × 4 × 3
= 60
So, 3 students can be arranged from 5 students in 60 ways.
Permutation of All Objects
If all n objects are arranged, then:
nPn = n!
Example:
Number of ways of arranging 5 books on a shelf:
5! = 5 × 4 × 3 × 2 × 1 = 120
So, 5 books can be arranged in 120 ways.
Permutation When Repetition is Not Allowed
When repetition is not allowed, an object used once cannot be used again.
Example:
How many 3-digit numbers can be formed using 1, 2, 3, 4 without repetition?
Hundreds place = 4 choices
Tens place = 3 choices
Units place = 2 choices
Total numbers = 4 × 3 × 2 = 24
This can also be written as:
4P3 = 4!/(4 - 3)! = 24
Word Arrangements
Word arrangement questions are common in Permutations and Combinations Class 11 Notes.
Example:
How many words can be formed using all letters of the word MATH?
There are 4 distinct letters.
Number of arrangements = 4!
= 4 × 3 × 2 × 1
= 24
So, 24 arrangements can be formed.
Number Formation Using Permutations
Permutation is used when digits are arranged to form numbers.
Example:
How many 3-digit numbers can be formed using 0, 2, 3, 5, 7 without repetition, if the number is greater than 300?
Hundreds place can be 3, 5 or 7.
So, hundreds place = 3 choices.
After that, 4 digits remain for tens and units place.
Tens and units place can be filled in:
4P2 = 4!/2! = 12 ways
Total numbers = 3 × 12 = 36
So, 36 numbers can be formed.
What is a Combination?
A combination is a selection of objects where order does not matter.
Example:
Suppose A, B and C are three students.
Selecting A and B is the same as selecting B and A.
So, AB and BA are not counted separately.
This is a combination problem.
Combination Formula
The number of combinations of n different objects taken r at a time is:
nCr = n!/[r!(n - r)!]
Here:
n = total number of objects
r = number of objects selected
Condition:
0 ≤ r ≤ n
Example:
Find the number of ways of selecting 2 students from 8 students.
n = 8
r = 2
8C2 = 8!/[2!(8 - 2)!]
= 8!/(2!6!)
= (8 × 7)/2
= 28
So, 2 students can be selected from 8 students in 28 ways.
Permutation vs Combination
The main difference is order.
| Basis | Permutation | Combination |
| Meaning | Arrangement | Selection |
| Order | Order matters | Order does not matter |
| Formula | nPr = n!/(n - r)! | nCr = n!/[r!(n - r)!] |
| Example | Arranging students in a row | Selecting students for a team |
| Counts AB and BA | Different | Same |
If the question asks for arranging, ranking, seating or forming numbers, use permutation.
If the question asks for selecting, choosing, forming a group or committee, use combination.
Relation Between Permutation and Combination
Permutation and combination are connected.
The relation is:
nPr = nCr × r!
This means:
First select r objects from n objects.
Then arrange those r selected objects.
So:
Arrangement = Selection × Arrangement of selected objects
Example:
Select 3 students from 5 and arrange them.
Selection = 5C3
Arrangement of selected students = 3!
Total arrangements = 5C3 × 3! = 5P3
Important Properties of nCr
The main properties of combinations are:
| Property | Meaning |
| nC0 = 1 | Selecting nothing from n objects |
| nCn = 1 | Selecting all n objects |
| nC1 = n | Selecting one object from n objects |
| nCr = nC(n-r) | Selecting r is same as leaving n - r |
| nCr + nC(r-1) = (n+1)Cr | Useful identity in combinations |
Example:
10C3 = 10C7
This is because selecting 3 objects from 10 is the same as leaving 7 objects.
When to Use nPr and nCr
| Question Type | Use |
| Arranging books | nPr |
| Forming numbers | nPr |
| Seating people in a row | nPr |
| Ranking students | nPr |
| Selecting students | nCr |
| Forming a committee | nCr |
| Choosing cards | nCr |
| Selecting points to form a triangle | nCr |
This table helps students choose the correct formula quickly.
Restrictions in Counting
Some counting questions include restrictions.
Examples:
- A number must be greater than 300.
- Repetition is not allowed.
- A committee must include 1 boy and 3 girls.
- Some objects must always be together.
- Some objects cannot be placed together.
In such questions, read the condition first.
Then decide the number of choices for each step.
Committee Selection
Combination is used in committee selection because order does not matter.
Example:
How many committees of 4 students can be formed from 10 students?
10C4 = 10!/[4!6!]
= (10 × 9 × 8 × 7)/(4 × 3 × 2 × 1)
= 210
So, 210 committees can be formed.
Selection With Categories
When selection is made from different groups, use combination with multiplication.
Example:
How many groups can be formed by selecting 1 boy and 3 girls from 5 boys and 5 girls?
Number of ways to select 1 boy = 5C1 = 5
Number of ways to select 3 girls = 5C3 = 10
Total ways = 5 × 10 = 50
So, 50 groups can be formed.
Geometrical Applications of Combinations
Combinations are also used in geometry.
Number of Lines
Two points determine one straight line.
So, if there are n points and no three are collinear, the number of lines is:
nC2
Example:
Number of lines from 8 points:
8C2 = 28
Number of Triangles
Three non-collinear points determine one triangle.
So, if there are n points and no three are collinear, the number of triangles is:
nC3
Example:
Number of triangles from 8 points on a circle:
8C3 = 56
Difference Between Arrangement and Selection
| Situation | Arrangement or Selection |
| Form a 3-digit number | Arrangement |
| Select 3 players | Selection |
| Arrange 5 books | Arrangement |
| Choose 2 cards | Selection |
| Make a password | Arrangement |
| Form a committee | Selection |
A quick test:
Ask, “Will changing the order create a new answer?”
If yes, it is permutation.
If no, it is combination.
Important Formulas in Class 11 Maths Permutations and Combinations Notes
| Concept | Formula |
| Factorial | n! = n × (n - 1) × ... × 1 |
| Permutation | nPr = n!/(n - r)! |
| Combination | nCr = n!/[r!(n - r)!] |
| Relation | nPr = nCr × r! |
| All arrangements | nPn = n! |
| Basic combination property | nCr = nC(n-r) |
| Selection of all objects | nCn = 1 |
| Selection of no object | nC0 = 1 |
Solved Examples on Permutations and Combinations
Example 1: Find 5P2
Solution:
5P2 = 5!/(5 - 2)!
= 5!/3!
= 5 × 4
= 20
So, 5P2 = 20.
Example 2: Find 8C2
Solution:
8C2 = 8!/[2!(8 - 2)!]
= 8!/(2!6!)
= (8 × 7)/(2 × 1)
= 28
So, 8C2 = 28.
Example 3: Arrange the Letters of ROCKY
How many four-letter words can be formed using the letters of ROCKY without repetition?
The word ROCKY has 5 distinct letters.
We need to arrange 4 letters at a time.
Number of words = 5P4
= 5!/(5 - 4)!
= 5!/1!
= 120
So, 120 four-letter words can be formed.
Example 4: Select Students
How many ways can 2 students be selected from 8 students?
Since order does not matter, use combination.
8C2 = 8!/[2!6!]
= 28
So, 2 students can be selected in 28 ways.
Example 5: Find Number of Diagonals in an Octagon
An octagon has 8 vertices.
Number of line segments formed by joining any two vertices:
8C2 = 28
Out of these, 8 are sides.
Number of diagonals = 28 - 8 = 20
So, an octagon has 20 diagonals.
Common Mistakes in Permutations and Combinations
| Mistake | Correct Approach |
| Using nPr for selection | Use nCr when order does not matter |
| Using nCr for arrangement | Use nPr when order matters |
| Forgetting restrictions | Read conditions before applying formula |
| Treating AB and BA the same in arrangement | Count them separately in permutations |
| Treating AB and BA as different in selection | Count them once in combinations |
| Expanding factorials fully every time | Cancel common factors first |
Quick Highlights of CBSE Class 11 Maths Notes Chapter 6
| Topic | Quick Revision Point |
| Counting | Finding number of possible outcomes |
| Addition principle | Used for separate cases |
| Multiplication principle | Used for step-by-step choices |
| Factorial | Product of consecutive positive integers |
| Permutation | Arrangement where order matters |
| Combination | Selection where order does not matter |
| nPr formula | nPr = n!/(n - r)! |
| nCr formula | nCr = n!/[r!(n - r)!] |
| Word arrangement | Usually uses permutation |
| Number formation | Usually uses permutation |
| Committee selection | Usually uses combination |
| Geometry questions | Lines use nC2, triangles use nC3 |
Important Terms from CBSE Class 11 Maths Revision Notes Chapter 6
The terms below cover the main definitions students need while revising this chapter.
| Term | Meaning |
| Counting | Finding the number of possible outcomes |
| Fundamental principle of counting | Rule used to count outcomes systematically |
| Addition principle | Counting rule used when one case or another case occurs |
| Multiplication principle | Counting rule used when choices happen in steps |
| Factorial notation | Product written as n! |
| Permutation | Arrangement of objects where order matters |
| Combination | Selection of objects where order does not matter |
| nPr | Number of permutations of n objects taken r at a time |
| nCr | Number of combinations of n objects taken r at a time |
| Arrangement | Ordered placement of objects |
| Selection | Choosing objects without considering order |
| Restriction | Extra condition given in a counting question |
| Word arrangement | Arrangement of letters of a word |
| Number formation | Forming numbers from given digits |
| Committee selection | Choosing people from a group |
Useful Links for Class 11 Maths
| Section | Useful Links |
| Syllabus | CBSE Class 11 Maths Syllabus |
| Revision Notes | CBSE Class 11 Maths Revision Notes |
| Maths Notes | CBSE Class 11 Maths Revision Notes Chapter 1 |
| Maths Notes | CBSE Class 11 Maths Revision Notes Chapter 2 |
| NCERT Solutions | NCERT Solutions Class 11 Maths |
| Sample Papers | CBSE Sample Papers for Class 11 Maths |
| Important Questions | Important Questions Class 11 Maths |
| NCERT Books | NCERT Books for Class 11 Maths |
Q.1 A man wants to cut three lengths from a single piece of board of length 91cm. The second length is to be 3cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5cm longer than the second?
Ans
Let x be the length of the shortest board,
then (x + 3) and 2x are the lengths of the second and third piece, respectively
Thus x + (x + 3) + 2x ≤ 91
⇒ 4x + 3 ≤ 91
⇒ 4x ≤ 88
⇒ x ≤ 22
According to the problem,
2x ≥ (x +3) + 5
⇒ x ≥ 8
The possible lengths of the shortest board are greater than or equal to 8 but less than or equal to 22.
Q.2 A solution is to be kept between 68° F and 77° F. What is the range of temperature in degree Celsius (C) ? if the Celsius(C) / Fahrenheit (F) conversion formula is given by F = (9/5)C + 32.
Ans
We have, F = (9/5)C + 32 and 68° < F° < 77°
Therefore, 68° < (9/5) C + 32° < 77°
Subtracting 32 from each side, we get
68° – 32° < (9/2) C < 77° – 32°
36° < (9/5) C < 45°
Multiply by (5/9) we get
36° × (5/9) < C < 45° × (5/8)
20° < C < 25°.
Hence required range lies between 20°C and 25°C.
Q.3 How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 20% but less than 30% acid content?
Ans
Let x litres water is added to 45% solution of acid, therefore,
20% of (1125 + x) < (1125) × 45/100 or (1125 + x)/5 < (1125) × 45/100
Multiplying by 20, we get
4(1125 + x) < 10125 or
4500 + 4x < 10125
Subtracting 4500 from both sides, we get
4x < 10125 – 4500 = 5625 or
x < 1406.25 …(1)
Now,
30% of (1125 + x) > (1125) × 45/100 or 3(1125 + x)/10 > (1125) × 45/100
Multiplying by 20, we get
6(1125 + x) > 10125 or 6750 + 6x > 10125
Subtracting 6750 from both sides, we get
6x > 10124 – 6750 = – 3375 or
x > 562.5 …(2)
Therefore, from (1) and (2), we get
1406.25 < x < 900.
Q.4 A manufacturer has 600 litres of a 10% solution of acid. How many litres of a 20% acid solution must be added to it so that acid content in the resulting mixture will be more than 15% but less than 18%?
Ans
Let x litres of 20% acid solution is required to be added. Then Total mixture = (x + 600) litres
Therefore,
20% × + 10% of 600 > 15% of (x + 600) and
20% × + 10% of 600 < 18% of (x + 600 or
(20/100)x + (10/100) × 600 > (15/100) (x + 600) and
(20/100)x + (10/100) × 600 < (18/100) (x + 600) or
20x + 6000 > 15x + 9000 and
20x +6000 < 18x + 10800 or
5x > 3000 and 2x < 4800 or
x > 600 and x < 2400,
i.e., 600 < x < 2400.
Thus, the number of litres of the 20% solution of acid will have to be more than 600 litres but less than 2400 litres.
Q.5 Solve the following system of inequalities graphically 3x + 2y ≤ 150, x + 4y ≤ 150, x ≤ 15, x, y ≥ 0.
Ans
We first draw the graph of the lines corresponding to given inequalities.
To draw the graph of line, we need at least two solutions.
Two solution for the line 3x + 2y = 150 are:
| x | 0 | 50 |
| y | 75 | 0 |
Two solutions for the line x + 4y = 80 are:
| x | 0 | 80 |
| y | 20 | 0 |
Given x ≤ 15
x = 15 is a line parallel to y-axis and 15 units apart from the origin in the positive direction of x-axis.
Since, x ≥ 0, y ≥ 0 the solution region lies only in the first quadrant.
Thus graphical representation of the required solution is given in figure by the shaded region.
Q.6 Solve the following system of inequalities graphically 4x + 3y ≤ 60, y ≥ 2x, x ≥ 3 where x, y ≥ 0.
Ans
We first draw the graph of the lines corresponding to given inequalities.
To draw the graph of line, we need at least two solutions
Two solutions for the line 4x + 3y = 60 are:
| x | 0 | 15 |
| y | 20 | 0 |
Two solutions for the line y = 2x are:
| x | 0 | 5 |
| y | 0 | 10 |
Given x ≥ 3, x = 3 is a line parallel to y-axis and 3 units apart from origin in the positive direction of x-axis.
Since, x ≥ 0, y ≥ 0. So the solution region lies only in the first quadrant.
Thus graphical representation of the required solution is given in figure by shaded region.
Q.7 Solve the following system of inequalities graphically:
5x + 4y ≤ 40 …(1)
x ≥ 2 …(2)
y ≥ 3 …(3)
Ans
To draw the graph of line, we need at least two solutions.
Two solutions for the line 5x + 4y = 40 are:
| x | 0 | 8 |
| y | 10 | 0 |
Given x ≥ 2 and y ≥ 3
x = 2 is a line parallel to the y-axis and 2 units apart from the origin in the positive direction of x-axis.
y = 3 is a line parallel to the x-axis and 3 units apart from the origin in the positive direction of y-axis.
Since, x ≥ 0, y ≥ 0. So, the solution region lies only in the first quadrant.
Thus graphical representation of the solutions are given in Fig. by shaded region.
Q.8 Solve the following system of inequalities graphically:
x + 2y ≤ 8 …(1)
2x + y ≤ 8 …(2)
x ≥ 0 …(3)
y ≥ 0 …(4)
Ans
To draw the graph of line, we need at least two solutions.
Two solutions for the line x + 2y = 8 are:
| x | 0 | 8 |
| y | 4 | 0 |
and two solution for the line 2x + y = 8 are:
| x | 0 | 4 |
| y | 8 | 0 |
Since x ≥ 0, y ≥ 0, therefore solutions lie in first quadrant.
Thus graphical representation of the solutions are given in figure and every point in the shaded region represents a solution of the given system of inequalities.
Q.9 Solve the following system of inequalities graphically:
2x + y ≥ 6, 3x + 4y < 12.
Ans
To draw the graph of line, we need at least two solutions.
Two solutions for the line 2x + y = 6 …(1)
| x | 0 | 3 |
| y | 6 | 0 |
and two solutions for the line 3x + 4y = 12 …(2)
| x | 0 | 4 |
| y | 3 | 0 |
Thus graphical representation of the solutions are given in figure by shaded region.
Q.10 Solve the system of inequalities:
3x – 7 < 5 + x …(1)
11 – 5x ≤ 1 …(2)
and represent the solutions on the number line.
Ans
From inequality (1), we have
3x – 7 < 5 + x …(1)
or x < 6 …(3)
Also, from inequality (2), we have
11 – 5x ≤ 1 …(2)
– 5x ≤ – 10, .i.e., x ≥ 2 …(4)
We draw the graph of inequalities (3) and (4) on the number line, the values of x, which are common to both,
are shown by a bold line on the number line in figure.
Thus, solutions of the system are real numbers x lying between 2 and 6 including 2, i.e., 2 ≤ x < 6.
Q.11 The longest side of a triangle is 4 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.
Ans
Let AB be the shortest side of the triangle = x cm
Then the longest side will be = 4x cm
And the third side will be = 4x – 2
Perimeter of the triangle is at least 61 cm.
x + 4x + 4x – 2 ≥ 61
⇒ 9x – 2 ≥ 61
⇒ x ≥ 7.
Hence, the minimum length of the shortest side is 7 cm.
Q.12 Solve the inequality 3y – 5x < 30 graphically in two-dimensional plane.
Ans
To draw the graph of line 3y – 5x = 30, we need at least two solutions which are:
| x | 0 | – 6 |
| y | 10 | 0 |
Thus graphical representation of the solutions are given in figure by shaded region.
Q.13 Solve the inequality y + 8 ≥ 2x graphically in two-dimensional plane.
Ans
To draw the graph of line y + 8 = 2x, we need at least two solutions which are:
| x | 0 | 4 |
| y | – 8 | 0 |
Thus graphical representation of the solutions are given in figure by the shaded region.
Q.14 Solve 7x + 3 < 5x + 9. Show the graph of the solutions on number line.
Ans
We have 7x + 3 < 5x + 9
⇒ 2x < 6 or x < 3
The graphical representation of the solutions are given in figure
Q.15
Ans
Q.16 Solve 4x + 3 < 6x + 7.
Ans
We have, 4x + 3 < 6x + 7
⇒ 4x – 6x + 3 < 6x – 6x + 7
⇒ –2x + 3 < 7
⇒ –2x < 7 – 3
⇒ –2x < 4
⇒ x > –2
i.e., all the real numbers which are greater than –2, are the solutions of the given inequality.
Hence, the solution set is (–2, ∞).
Q.17
Ans
Q.18 Solve (x – 3)/(x – 5) > 0.
Ans
Here, (x – 3)/(x – 5) > 0
x = 3, 5 are critical points
Hence from figure (x – 3)/(x – 5) > 0
⇒ x ∈ (– ∞, 3) ∪ (5, ∞).
Q.19 Solve 7x + 9 ≥ 30.
Ans
Here, 7x + 9 ≥ 30.
⇒ 7x ≥ 30 – 9
⇒ x ≥ 3
⇒ x ∈ [3, ∞).
Q.20 Solve 5x – 3 < 3x + 1, where x∈ N.
Ans
Here, 5x – 3 < 3x + 1
5x – 3x < 3 + 1
2x < 4
x < 2
Therefore if x is less than 2 then the value of x is 1.
Q.21 Draw the graph of |x| ≤ 3.
Ans
Q.22
Ans
On combining all the three conditions, we get the above shaded region, which represents solution of system of inequalities.
Q.23
Ans
Q.24
Ans
Q.25
Ans
Q.26
Ans
Q.27
Ans
Q.28 The marks obtained by Beeru of class XI in first and second terminal examination are 70 and 45 respectively. Find the number of minimum marks he should get in the annual examination to have an average of atleast 60 marks.
Ans
Q.29
Ans
Thus, here the dark and thick line represent solution of linear inequality.
Q.30
Ans
We have,
37 – 5x < 17
⇒ –5x < 17 – 37
⇒ –5x < –20
Divide both sides by –5,
⇒ x > 4
Since, on dividing by a – ve number, the sign of inequality changes,
⇒ 4 < × > ∞
⇒ x ∈ (4, ∞)
Q.31 Find the integral value of x: 5x – 5 < 3x + 1.
Ans
We have, 5x – 5 < 3x + 1
⇒ 5x – 3x < 1 + 5
⇒ 2x < 6
⇒ x < 3
⇒ x = –∞ … –3, –2, –1, 0, 1, 2 = (–∞, 2]
Q.32
Ans
FAQs (Frequently Asked Questions)
Permutations and Combinations is a chapter about counting. Permutations deal with arrangements where order matters. Combinations deal with selections where order does not matter.
In permutation, order matters. In combination, order does not matter. Arranging three students in a row is a permutation. Selecting three students for a team is a combination.
The permutation formula is nPr = n!/(n – r)!. It gives the number of ways of arranging r objects from n different objects.
The combination formula is nCr = n!/[r!(n – r)!]. It gives the number of ways of selecting r objects from n different objects.
Use nPr when order matters, such as arrangements, rankings or number formation. Use nCr when order does not matter, such as selection, committee formation or choosing cards.