Class 11 Mathematics (041) · Chapter 6 NotesPermutations and Combinations
Learn about permutations and combinations in Class 11 Mathematics (041). Understand the fundamental principle of counting, permutations of distinct and non-distinct…
This chapter introduces fundamental concepts in combinatorics, focusing on counting techniques for arrangements (permutations) and selections (combinations) of objects. You'll learn to calculate the number of ways to arrange or select items under different conditions, such as with or without repetition, and with or without distinct objects. These techniques are essential for solving problems in probability, algebra, and various real-world scenarios involving arrangements and selections.
What you'll learn
1Understand the fundamental principle of counting and apply it to solve problems
2Calculate permutations of distinct objects with and without repetition
3Determine the number of permutations when not all objects are distinct
4Compute combinations and understand the relationship between permutations and combinations
5Apply factorial notation and its properties in counting problems
6Solve practical problems using the formulas for permutations and combinations
Chapter at a glance
01Fundamental Principle of Counting
02Permutations
03Permutations when all objects are distinct
04Permutations when repetition is allowed
05Permutations when not all objects are distinct
06Combinations
07Relationship between Permutations and Combinations
Detailed chapter notes
01
Fundamental Principle of Counting
The fundamental principle of counting, also known as the multiplication principle, states that if one event can occur in m different ways and a second event can occur independently in n different ways, then the two events can occur in m × n different ways. This principle can be extended to any number of events. For example, if you have 3 pants and 2 shirts, you can dress up in 3 × 2 = 6 different pairs. This principle is foundational for solving more complex counting problems.
If event A can occur in m ways and event B can occur in n ways, the total number of ways both events can occur is m × n
The principle can be extended to more than two events
02
Permutations
Permutations are arrangements of objects in a specific order. The number of permutations of n distinct objects taken r at a time is denoted by nPr and is calculated using the formula nPr = n! / (n-r)!, where n! (n factorial) is the product of all positive integers up to n. This formula helps in determining the number of ways to arrange objects when the order matters.
When all objects are distinct and repetition is not allowed, the number of permutations is calculated using the formula nPr = n! / (n-r)!. For example, the number of 3-digit numbers that can be formed from the digits 1, 2, 3, 4, and 5 without repetition is 5P3 = 5! / (5-3)! = 60. This formula is widely used in problems involving arrangements where each object is unique and used only once.
Permutations of distinct objects without repetition: nPr = n! / (n-r)!
Example5P3 = 60
04
Permutations with Repetition
When repetition of objects is allowed, the number of permutations is calculated using the formula nPr = n^r. For example, the number of 3-digit numbers that can be formed from the digits 1, 2, 3, 4, and 5 with repetition allowed is 5^3 = 125. This formula is useful in problems where objects can be used more than once in different positions.
Permutations with repetitionnPr = n^r
Example5^3 = 125
05
Permutations of Non-Distinct Objects
When not all objects are distinct, the number of permutations is adjusted by dividing by the factorial of the count of each repeated object. For example, the number of distinct arrangements of the letters in the word 'ROOT' is 4! / 2! = 12, because the letter 'O' is repeated twice. This adjustment accounts for the indistinguishable arrangements caused by repeated objects.
Combinations are selections of objects where the order does not matter. The number of combinations of n distinct objects taken r at a time is denoted by nCr and is calculated using the formula nCr = n! / (r! × (n-r)!). This formula is essential for problems involving selections, such as choosing a committee from a group of people, where the order of selection is irrelevant.
Combination formulanCr = n! / (r! × (n-r)!)
Special casenC0 = 1
07
Relationship Between Permutations and Combinations
There is a direct relationship between permutations and combinations, expressed by the formula nPr = nCr × r!. This relationship helps in converting between permutations and combinations when the order of selection becomes relevant. For example, if you need to find the number of ways to arrange 3 selected items out of 5, you can use the combination formula to find the number of ways to choose 3 items and then multiply by the number of ways to arrange them.
RelationshipnPr = nCr × r!
Example5P3 = 5C3 × 3! = 10 × 6 = 60
Want the complete chapter resources?Topic notes, quizzes and flashcards for Permutations and Combinations.
In how many ways can a person choose one item from a menu with 5 appetizers, 4 main courses, and 3 desserts?
A12
B20
C60
D5 + 4 + 3
Show answer
Answer: (A) 12
By the fundamental principle of counting, the total number of ways to choose one item is 5 + 4 + 3 = 12, since the person must choose exactly one from one category.
Question 02
The number of permutations of the letters of the word 'BOOK' is:
A24
B12
C6
D4
Show answer
Answer: (B) 12
The word 'BOOK' has 4 letters with O repeated twice. Number of permutations = 4!/2! = 24/2 = 12.
Question 03
In how many ways can the letters of the word 'BOOK' be arranged?
A24
B12
C6
D4
Show answer
Answer: (B) 12
The word 'BOOK' has 4 letters with O repeated twice. Number of arrangements = 4!/2! = 24/2 = 12.
Question 04
How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5 if repetition of digits is allowed?
A75
B125
C60
D100
Show answer
Answer: (B) 125
When repetition is allowed, each of the 3 positions can be filled in 5 ways. Total = 5 × 5 × 5 = 125.
Question 05
How many distinct permutations can be formed using the letters of the word 'BOOK'?
A24
B12
C6
D8
Show answer
Answer: (B) 12
The word 'BOOK' has 4 letters with O repeated twice. Using the formula n!/n₁!n₂!, we get 4!/(2!) = 24/2 = 12 distinct permutations.
Ready for more practice?Unlock the full quiz for this chapter.
What is the difference between permutations and combinations?
Permutations are arrangements where the order of selection matters, while combinations are selections where the order does not matter. For example, the arrangement ABC is different from CBA in permutations, but both are the same combination.
When should I use the permutation formula with repetition?
Use the permutation formula with repetition (nPr = n^r) when the same object can be used more than once in different positions. For example, forming numbers where digits can be repeated.
How do I handle repeated objects in permutations?
When objects are repeated, divide the total permutations by the factorial of the count of each repeated object. For example, the word 'ROOT' has 4! / 2! = 12 distinct arrangements because the letter 'O' is repeated twice.
What is the relationship between permutations and combinations?
The relationship is given by nPr = nCr × r!. This means the number of permutations is equal to the number of combinations multiplied by the number of ways to arrange the selected items.
Why is 0! defined as 1?
0! is defined as 1 to maintain consistency in mathematical formulas and to provide a base case for recursive definitions of factorials. It represents the number of ways to arrange zero objects, which is logically one way (doing nothing).
How do I choose between using permutations and combinations in a problem?
Use permutations when the order of selection matters, such as arranging people in a line. Use combinations when the order does not matter, such as selecting a committee from a group of people.