Problem 47 - Algebra and Trigonometry

Question

These problems involve combinations. Committee In how many ways can a committee of three members be chosen from a club of 25 members?

Step-by-Step Solution

Answer

The committee can be chosen in 2300 ways.

Step 1: Understand the Problem

To solve this problem, we need to determine how many ways we can choose a committee of 3 members from a total of 25 members. This is a classical combination problem, where the order of selection does not matter.

Step 2: Formula for Combinations

The formula for combinations is given by \( C(n, k) = \frac{n!}{k!(n-k)!} \), where \( n \) is the total number of items to choose from, and \( k \) is the number of items to choose. Here, \( n = 25 \) and \( k = 3 \).

Step 3: Substitute Values into the Formula

Substitute the values into the combination formula: \[ C(25, 3) = \frac{25!}{3!(25-3)!} = \frac{25!}{3! \times 22!} \]

Step 4: Simplify the Factorials

Simplify the factorial expression by canceling out the common terms in numerator and denominator: \[ C(25, 3) = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} \].

Step 5: Compute the Result

Compute the result of the expression: \( \frac{25 \times 24 \times 23}{6} = 2300 \).

Key Concepts

Understanding Factorials

Factorials are a fundamental concept in combinatorics. The factorial of a number, denoted as \( n! \), is the product of all positive integers less than or equal to \( n \). For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

Factorials are essential when calculating combinations and permutations. They help in determining the total number of ways items can be arranged or selected.

• Factorial of \( n \), \( n! \), is used to find the total arrangements.
• In combination problems, factorials help to ignore the order of selection.

When dealing with large numbers, factorials grow very quickly, which makes calculations complex if not properly simplified.

In the example given, we simplified the factorial calculation by canceling out common terms in the numerator and denominator. This makes calculations manageable and is a crucial step when working with combination formulas.

Choosing a Committee with Combinations

A committee selection problem is a classic example of how combinations are applied. It involves selecting a specific number of members from a larger group. The main idea is that the order in which members are selected doesn't matter.

To choose a committee, you use the concept of combinations, which groups items without regard to order. Combinations are denoted by \( C(n, k) \), where \( n \) is the total number of items, and \( k \) is the number of items to choose.

In practical terms:

• Consider a club with 25 members.
• You want to form a committee of 3 members.

You calculate how many ways you can select these committee members by considering all possible groupings, not sequences, thus applying combination formulas.

Using the Combination Formula

The combination formula \( C(n, k) = \frac{n!}{k!(n-k)!} \) is used to determine the number of ways to choose \( k \) items from \( n \) items without considering the order of selection.

To use this formula effectively:

• Identify the total number \( n \) and the selection number \( k \).
• Calculate the factorial of \( n \), \( k \), and \( n-k \).
• Plug these values into the formula.

For example, choosing 3 members from a group of 25 involves computing \( C(25, 3) \). We follow these steps:

• Calculate \( 25! \), but focus on the top terms needed.
• Simplify by computing \( \frac{25 \times 24 \times 23}{3!} \), considering cancellations.
• This results in the total number of combinations, 2300, showing all possible groups.