Introduction to Permutation and Combination
Permutation and Combination are crucial topics in the branch of mathematics known as combinatorics. They deal with counting, arranging, and selecting objects from a given set. These concepts are widely applied in probability, statistics, and real-life situations where arrangements or selections need to be made.
1. Multiplication Principle: - If one task can be done in m ways and another task can be done in n ways, then the two tasks together can be done in m×n ways.
2. Permutation: - Permutation refers to the arrangement of objects in a specific order. The order of the arrangement matters in permutations. If you're asked to arrange objects or people in different orders, you're dealing with permutations.
Key Formula for Permutation: - The number of ways to arrange r objects from a set of n distinct objects is given by: - P(n, r) = n!/(n - r)
Where n! (n factorial or factorial n) is the product of all integers from 1 to n.
i.e n! = n×(n-1)×(n-2)×(n-3)×.......... ×2×1.
3. Combination: - Combination refers to the selection of objects where the order does not matter. If you're selecting objects or people without regard to the order, you're dealing with combinations.
Key Formula for Combination: - The number of ways to select r objects from a set of n distinct objects is given by: - C(n, r) = n!/r!(n - r)!
In this formula, the order of selection is irrelevant.
4. Important Concepts: - Factorial n!: - Factorial of a number n, denoted n!, is the product of all positive integers up to n.
i.e n! = n×(n-1)×(n-2)×(n-3)×.......... ×2×1.
5. Summary of Formulas: -
(i) Permutation Formula: - P(n, r) = n!/(n - r)!
(ii) Combination Formula: - C(n, r) = n!/r!(n - r)!
6. Special Case: - When r = n,
P(n, n) = n!
C(n, n) = 1
7. Example Problems: -
(i) Permutation Example: - How many ways can 3 students be arranged in a line from a group of 5?
P(5, 3) = 5!/(5-3)! = 5!/2! = 120/2 = 60
(ii) Combination Example: - How many ways can 3 students be selected from a group of 5?
C(5, 3) = 5!/3!(5-3)! = 5!/3!×2 = 120/6×2 = 10
Understanding these concepts and formulas is crucial for solving problems related to arrangements and selections in mathematics.
Problems based on multiplication Principle
Problems on Multiplication Principle
Problems based on Permutation
Problems on Permutation
Problems based on Combination
Problems on Combinations
Difference between permutation and combination
What is permutation?
What is Combination?

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