10 Essential Problems on Permutation and Combination

Understanding Permutation and Combination: 10 Essential Problems

Permutation and combination are fundamental concepts in mathematics, and they have numerous applications in various fields, including computer science, statistics, and engineering. In this article, we will explore 10 essential problems on permutation and combination, along with their solutions and explanations.

Problem 1: Permutation of Objects

A box contains 5 different objects: A, B, C, D, and E. How many different ways can these objects be arranged in a row?

Solution: The number of permutations of n objects is given by n!. In this case, n = 5, so the number of permutations is 5! = 5 × 4 × 3 × 2 × 1 = 120.

Problem 2: Combination of Objects

A committee of 3 members is to be formed from a group of 7 people. How many different combinations of 3 members can be formed?

Solution: The number of combinations of n objects taken r at a time is given by the formula nCr = n! / (r!(n-r)!). In this case, n = 7 and r = 3, so the number of combinations is 7C3 = 7! / (3!(7-3)!) = 35.

Problem 3: Permutation with Repetition

A password consists of 4 characters, and each character can be any of the 26 letters of the alphabet. How many different passwords can be formed if repetition is allowed?

Solution: Since repetition is allowed, the number of permutations is given by the formula n^r, where n is the number of options and r is the number of selections. In this case, n = 26 and r = 4, so the number of permutations is 26^4 = 456,976.

Problem 4: Combination with Repetition

A bakery sells 5 different types of cakes. How many different combinations of 3 cakes can be formed if repetition is allowed?

Solution: Since repetition is allowed, the number of combinations is given by the formula (n+r-1)! / (r!(n-1)!). In this case, n = 5 and r = 3, so the number of combinations is (5+3-1)! / (3!(5-1)!) = 35.

Problem 5: Permutation of a Multiset

A bag contains 3 red balls, 2 blue balls, and 1 green ball. How many different arrangements of these balls can be formed?

Solution: The number of permutations of a multiset is given by the formula n! / (n1! × n2! ×… × nk!), where n is the total number of objects and n1, n2,…, nk are the numbers of each type of object. In this case, n = 6, n1 = 3, n2 = 2, and n3 = 1, so the number of permutations is 6! / (3! × 2! × 1!) = 60.

Problem 6: Combination of a Multiset

A bag contains 3 red balls, 2 blue balls, and 1 green ball. How many different combinations of 3 balls can be formed?

Solution: The number of combinations of a multiset is given by the formula n! / (r!(n-r)! × n1! × n2! ×… × nk!), where n is the total number of objects, r is the number of selections, and n1, n2,…, nk are the numbers of each type of object. In this case, n = 6, r = 3, n1 = 3, n2 = 2, and n3 = 1, so the number of combinations is 6! / (3!(6-3)! × 3! × 2! × 1!) = 20.

Problem 7: Circular Permutation

A circular table has 5 chairs. How many different arrangements of 5 people can be formed around the table?

Solution: The number of circular permutations is given by the formula (n-1)!. In this case, n = 5, so the number of permutations is (5-1)! = 4! = 24.

Problem 8: Permutation with Restriction

A bookshelf has 5 books, and 2 of them are identical. How many different arrangements of the books can be formed if the identical books must be together?

Solution: The number of permutations with restriction is given by the formula n! / (k1! × k2! ×… × km!), where n is the total number of objects and k1, k2,…, km are the numbers of identical objects. In this case, n = 5, and there are 2 identical books, so the number of permutations is 5! / (2! × 1! × 1! × 1!) = 60.

Problem 9: Combination with Restriction

A committee of 3 members is to be formed from a group of 7 people, and 2 of the members must be women. How many different combinations of 3 members can be formed?

Solution: The number of combinations with restriction is given by the formula n! / (r!(n-r)! × k1! × k2! ×… × km!), where n is the total number of objects, r is the number of selections, and k1, k2,…, km are the numbers of restricted objects. In this case, n = 7, r = 3, and there are 2 women, so the number of combinations is 7! / (3!(7-3)! × 2! × 1!) = 35.

Problem 10: Permutation and Combination Together

A quiz consists of 5 multiple-choice questions, each with 4 options. How many different ways can a student answer the questions if each question can be answered in only one way?

Solution: The number of permutations is given by the formula n^r, where n is the number of options and r is the number of selections. In this case, n = 4 and r = 5, so the number of permutations is 4^5 = 1024.

📝 Note: These problems cover a range of topics in permutation and combination, including permutations with repetition, combinations with repetition, permutations of a multiset, combinations of a multiset, circular permutations, permutations with restriction, combinations with restriction, and permutations and combinations together.

To summarize, permutation and combination are fundamental concepts in mathematics, and they have numerous applications in various fields. By mastering these concepts, you can solve a wide range of problems, from simple to complex. Remember to practice regularly and apply the formulas and techniques to real-world problems to become proficient in permutation and combination.