Problem 1 - Discrete Mathematics with Applications

Question

A committee consists of nine members. Find the number of subcommittees that can be formed of each size. Two

Step-by-Step Solution

Answer

For a committee of 9 members, the number of subcommittees that can be formed for each size are: - 1 member: 9 subcommittees - 2 members: 36 subcommittees - 3 members: 84 subcommittees - 4 members: 126 subcommittees - 5 members: 126 subcommittees - 6 members: 84 subcommittees - 7 members: 36 subcommittees - 8 members: 9 subcommittees - 9 members: 1 subcommittee

Step 1: Recall the combinations formula

The combinations formula is given by: \( C(n, k) = \frac{n!}{k!(n-k)!} \) Where: - C(n, k) represents the number of combinations of choosing k elements from a set of n elements - n! represents the factorial of n, which is the product of all positive integers less than or equal to n - k! represents the factorial of k, which is the product of all positive integers less than or equal to k

Step 2: Apply the combinations formula for each subcommittee size

We will apply the above formula to find the number of subcommittees that can be formed for each size from 1 to 9. For a subcommittee of size 1: \( C(9, 1) = \frac{9!}{1!(9-1)!} \) For a subcommittee of size 2: \( C(9, 2) = \frac{9!}{2!(9-2)!} \) For a subcommittee of size 3: \( C(9, 3) = \frac{9!}{3!(9-3)!} \) For a subcommittee of size 4: \( C(9, 4) = \frac{9!}{4!(9-4)!} \) For a subcommittee of size 5: \( C(9, 5) = \frac{9!}{5!(9-5)!} \) For a subcommittee of size 6: \( C(9, 6) = \frac{9!}{6!(9-6)!} \) For a subcommittee of size 7: \( C(9, 7) = \frac{9!}{7!(9-7)!} \) For a subcommittee of size 8: \( C(9, 8) = \frac{9!}{8!(9-8)!} \) For a subcommittee of size 9: \( C(9, 9) = \frac{9!}{9!(9-9)!} \)

Step 3: Calculate the number of subcommittees for each size

For a subcommittee of size 1: \( C(9, 1) = \frac{9!}{1!(9-1)!} = 9 \) subcommittees For a subcommittee of size 2: \( C(9, 2) = \frac{9!}{2!(9-2)!} = 36 \) subcommittees For a subcommittee of size 3: \( C(9, 3) = \frac{9!}{3!(9-3)!} = 84 \) subcommittees For a subcommittee of size 4: \( C(9, 4) = \frac{9!}{4!(9-4)!} = 126 \) subcommittees For a subcommittee of size 5: \( C(9, 5) = \frac{9!}{5!(9-5)!} = 126 \) subcommittees For a subcommittee of size 6: \( C(9, 6) = \frac{9!}{6!(9-6)!} = 84 \) subcommittees For a subcommittee of size 7: \( C(9, 7) = \frac{9!}{7!(9-7)!} = 36 \) subcommittees For a subcommittee of size 8: \( C(9, 8) = \frac{9!}{8!(9-8)!} = 9 \) subcommittees For a subcommittee of size 9: \( C(9, 9) = \frac{9!}{9!(9-9)!} = 1 \) subcommittee The number of subcommittees that can be formed for each size are: - 1 member: 9 subcommittees - 2 members: 36 subcommittees - 3 members: 84 subcommittees - 4 members: 126 subcommittees - 5 members: 126 subcommittees - 6 members: 84 subcommittees - 7 members: 36 subcommittees - 8 members: 9 subcommittees - 9 members: 1 subcommittee

Key Concepts

Combinatorial Analysis

Combinatorial analysis is a field of mathematics concerned with counting, enumeration, and finding structures in sets, usually finite. It encompasses various techniques and strategies to tally combinations in sets without having to list all possible variations, thus saving time and effort. For example, consider the problem of forming subcommittees from a group of nine members. Rather than listing every possible subset, combinatorial analysis provides methods like the combinations formula to calculate the number directly. This simplifies the process significantly and is incredibly useful in probability, statistics, and computer science, among other fields.

Problems solved by combinatorial analysis often involve organizing and arranging objects, choosing selections from a group, and deducing possibilities in games or puzzles. By understanding the basics of combinatorial analysis, students can tackle a wide range of problems more efficiently and with greater insight into the underlying mathematical principles.

Factorial Notation

Factorial notation is a fundamental mathematical concept, typically represented by an exclamation point (!), and is used to describe the product of an integer and all the positive integers below it. For instance, the factorial of 5, denoted as 5!, is calculated as 5 x 4 x 3 x 2 x 1, which equals 120. Factorials grow very rapidly with each additional number, making their manual calculation impractical for large numbers.

In combinatorial analysis, factorials are essential for formulas such as combinations and permutations because they help work out the total number of ways to arrange or select items. Practice in calculating smaller factorials and familiarity with properties, like the fact that 0! equals 1, can lead to a better grasp of more complex combinatorial problems. Despite its simplicity, the concept of factorials underpins many mathematical formulas and is a key to unlocking the world of discrete mathematics.

Binomial Coefficients

Binomial coefficients are the numbers that appear as coefficients in the expansion of a binomial power, for instance, in the expression \( (x + y)^n \). They are central to understanding the concept of combinations and can be found in Pascal's Triangle or calculated using the formula \( C(n, k) = \frac{n!}{k!(n-k)!} \). These coefficients tell us the number of ways to choose a set of elements from a larger set without considering order.

They play a pivotal role in probability and are frequently encountered in problems involving discrete random variables. The binomial theorem, which makes use of these coefficients, is a powerful tool that explains why the coefficients work the way they do in combinatorial settings. Learning to work with binomial coefficients often begins with simple problems, like counting the number of subcommittees from a larger committee before tackling more complex scenarios in statistics and algebra.