Chap 1. PERMUTATION AND COMBINATION

## Introduction Welcome to the exciting world of Permutation and Combination! In Additional Mathematics, we often deal with situations where we need to count the number of ways to arrange or select objects. This topic is fundamental to understanding probability and has wide applications in various fields, from computer science to genetics. This lesson will equip you with the tools to solve such counting problems systematically. ## Key Concepts ### Factorial Notation Factorial notation is a shorthand to represent the product of all positive integers up to a given integer. It is crucial for understanding permutations and combinations. * **Definition:** The factorial of a non-negative integer \( n \), denoted by \( n! \), is the product of all positive integers less than or equal to \( n \). * **Formula:** \( n! = n \times (n-1) \times (n-2) \times \dots \times 3 \times 2 \times 1 \) * **Special Case:** \( 0! = 1 \) (by definition, for consistency in formulas) * **Example:** * \( 3! = 3 \times 2 \times 1 = 6 \) * \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \) ### Permutation Permutation refers to the arrangement of objects in a specific order. The order of selection or arrangement matters significantly in permutations. #### Definition A permutation is an arrangement of all or part of a set of objects, where the order of the objects is important. #### Formula for Permutation of 'n' distinct objects taken 'r' at a time The number of permutations of \( n \) distinct objects taken \( r \) at a time is given by: \[ P(n,r) = rac{n!}{(n-r)!} \] Where: * \( n \) = total number of distinct objects available * \( r \) = number of objects to be arranged #### Special Cases of Permutations * **Permutation of \( n \) distinct objects taken all at a time:** When \( r = n \), the formula simplifies to \( P(n,n) = n! \). * **Permutation with Repeated Objects:** If there are \( n \) objects where \( n_1 \) are identical of one type, \( n_2 \) are identical of another type, ..., \( n_k \) are identical of a \( k \)-th type, the number of distinct permutations is: \[ P = rac{n!}{n_1! n_2! imes ext{...} imes n_k!} \] * **Circular Permutation:** The number of permutations of \( n \) distinct objects arranged in a circle is \( (n-1)! \). ### Combination Combination refers to the selection of objects where the order of selection does not matter. It's about forming groups or subsets. #### Definition A combination is a selection of all or part of a set of objects, where the order of the objects does not matter. #### Formula for Combination of 'n' distinct objects taken 'r' at a time The number of combinations of \( n \) distinct objects taken \( r \) at a time is given by: \[ C(n,r) = inom{n}{r} = rac{n!}{r!(n-r)!} \] Where: * \( n \) = total number of distinct objects available * \( r \) = number of objects to be selected #### Relationship between Permutation and Combination Since combination does not consider order, it can be derived from permutation by dividing by the number of ways to arrange the selected \( r \) objects: \( C(n,r) = rac{P(n,r)}{r!} \) #### Properties of Combination * \( C(n,r) = C(n, n-r) \) * \( C(n,0) = 1 \) * \( C(n,n) = 1 \) ### Distinguishing Between Permutation and Combination The key to solving problems involving arrangements or selections is to determine whether order matters. * **Order matters (Arrangement):** Use **Permutation** * Keywords: arrange, order, sequence, form a word, assign roles (e.g., president, vice-president) * **Order does NOT matter (Selection):** Use **Combination** * Keywords: select, choose, pick, form a group/committee, subset ## Vocabulary * **Permutation:** An arrangement of objects where order matters. * **Combination:** A selection of objects where order does not matter. * **Factorial:** The product of all positive integers less than or equal to a given positive integer (e.g., \( n! \)). * **Distinct objects:** Objects that are all different from each other. * **Arrangement:** The way in which things are placed or organized. * **Selection:** The act of choosing one or more items from a larger group. ## Examples ### Example 1: Factorial Calculation Calculate the value of \( 6! \). * **Solution:** \( 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \) ### Example 2: Simple Permutation In how many ways can 4 different books be arranged on a shelf? * **Analysis:** The order of the books matters. We are arranging all 4 books from a set of 4. This is a permutation of \( n \) objects taken \( n \) at a time. * **Solution:** \( P(4,4) = 4! = 4 \times 3 \times 2 \times 1 = 24 \) ways. ### Example 3: Permutation with Partial Selection How many different three-digit numbers can be formed using the digits 1, 2, 3, 4, 5 without repetition? * **Analysis:** The order of the digits matters (e.g., 123 is different from 321). We are selecting and arranging 3 digits from 5 distinct digits. * **Solution:** \( P(5,3) = rac{5!}{(5-3)!} = rac{5!}{2!} = rac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 5 \times 4 \times 3 = 60 \) numbers. ### Example 4: Permutation with Repeated Objects How many distinct arrangements can be made from the letters of the word "APPLE"? * **Analysis:** The word "APPLE" has 5 letters. The letter 'P' is repeated twice. So, \( n=5 \), \( n_P=2 \). * **Solution:** \( P = rac{5!}{2!} = rac{120}{2} = 60 \) distinct arrangements. ### Example 5: Simple Combination A committee of 3 students is to be chosen from a group of 8 students. In how many ways can the committee be formed? * **Analysis:** The order in which students are chosen for a committee does not matter. It's about who is in the group. This is a combination problem. * **Solution:** \( C(8,3) = rac{8!}{3!(8-3)!} = rac{8!}{3!5!} = rac{8 \times 7 \times 6}{3 \times 2 \times 1} = 8 \times 7 = 56 \) ways. ### Example 6: Combination with Multiple Groups A box contains 5 red balls and 4 blue balls. In how many ways can 2 red balls and 2 blue balls be selected? * **Analysis:** We need to select red balls AND blue balls. The selections are independent. Order does not matter for each selection. * **Solution:** * Ways to select 2 red balls from 5: \( C(5,2) = rac{5!}{2!3!} = rac{5 \times 4}{2 \times 1} = 10 \) * Ways to select 2 blue balls from 4: \( C(4,2) = rac{4!}{2!2!} = rac{4 \times 3}{2 \times 1} = 6 \) * Total ways = \( C(5,2) \times C(4,2) = 10 \times 6 = 60 \) ways. ## Exercises 1. Calculate \( 7! \). 2. Find the value of \( P(6,2) \). 3. Find the value of \( C(10,4) \). 4. In how many ways can a president, vice-president, and secretary be chosen from a club of 15 members? 5. How many different hands of 5 cards can be dealt from a standard deck of 52 cards? 6. How many distinct permutations can be formed from the letters of the word "MATHEMATICS"? ## Assignment 1. A football team consists of 11 players. If a captain and a vice-captain must be chosen, in how many ways can this be done? 2. From a group of 7 men and 6 women, a committee of 5 people is to be formed. Find the number of ways if: a. The committee consists of exactly 3 men and 2 women. b. The committee consists of at least 4 men. 3. How many 4-digit numbers can be formed from the digits 1, 2, 3, 4, 5, 6, 7 if no digit is repeated? 4. A student needs to answer 8 questions out of 10 questions in an examination. How many choices does the student have if: a. There are no restrictions. b. The first 3 questions are compulsory. ## Summary Today, we have explored the fundamental concepts of Permutation and Combination, essential tools for counting arrangements and selections. We learned about factorial notation (\( n! \)), which is the product of integers from 1 to \( n \). Permutations (\( P(n,r) \)) are used when the order of objects matters, while Combinations (\( C(n,r) \)) are used when the order does not matter. Remember to carefully analyze each problem to determine whether an arrangement or a selection is required, and then apply the appropriate formula. Mastery of these concepts will greatly assist you in solving more complex probability problems.