Problem 50 - Algebra and Trigonometry

Question

A four-person committee is to be elected from an organization’s membership of 11 people. How many different committees are possible?

Step-by-Step Solution

Answer

There are 330 different possible committees that can be elected.

Step 1: Understand Combination

A combination is a selection of items without considering the order. In this case, the order in which people are selected to make up the committee does not matter.

Step 2: Apply formula for combination

The formula for combinations is represented as \(C(n, r) = \frac{n!} {r!(n-r)!}\) where \(n\) is the total number of items available, \(r\) is the number of items to select, and \(!\) is the factorial operator indicating the product of an integer and all the integers below it, with \(0! = 1\).

Step 3: Substitute values into the formula

Here, we have \(n = 11\) (total members) and \(r = 4\) (committee members to be elected). So, \(C(11, 4) = \frac{11!}{4!(11-4)!}\).

Step 4: Calculate Factorials

First, compute the factorials: \(11! = 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \) and \(4! = 4 \times 3 \times 2 \times 1\), and \(7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\).

Step 5: Simplify

Now plug these factorials into the formula, so we get: \(C(11, 4) = \frac{11!}{4!(11-4)!} =\frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{ \(4 \times 3 \times 2 \times 1\) \times \(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\)}\). Next, cancel out the same terms from the numerator and the denominator. Hence, \(C(11, 4)= \frac{11 \times 10 \times 9 \times 8 }{ 4 \times 3 \times 2 \times 1}\).

Step 6: Calculate Result

Multiply the terms and divide to get the result: \(C(11, 4)= \frac{11 \times 10 \times 9 \times 8 }{ 4 \times 3 \times 2 \times 1} = 330\). So, there are 330 different possible committees.

Key Concepts

Factorial Calculation

Factorials are a fundamental concept in combinatorics, which help calculate the number of possible arrangements or selections. A factorial, represented by an exclamation mark (!), is the product of all positive integers up to a given number. For example,

• To calculate 5!, you'd multiply 5 by every positive integer less than it: \[5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\].
• For 7!, the calculation is \[7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040\].
• It's handy to remember that for any number \(n\), \(n!\) means multiplying all numbers from \(n\) down to 1.

Factorials are also used to define the concept of permutations and combinations, where order matters and does not matter, respectively. They provide the backbone for calculating large sets using formulas derived from factorial calculations.

Combinatorial Selection

Understanding how to make selections without regard to order is essential in cominuation math. Combinatorial selection, also known as combinations, helps determine how many ways you can choose a subset from a larger set.

• You use the combination formula, \( C(n, r) = \frac{n!}{r!(n-r)!} \), where \( n \) is the total items to choose from and \( r \) is the number of items to select.
• In our exercise, choosing 4 people out of 11 members doesn't require caring about the order of selection; hence, we use combinations rather than permutations.

For instance, if a committee needs to select 4 members out of 11, we'll substitute directly:

• By doing the factorial calculation, we simplify the formula to get: \[ C(11, 4) = \frac{11!}{4!7!} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330\].

This ensures all possible groupings of four out of eleven are considered, without duplication, resulting in 330 unique combinations.

Permutations and Combinations

One of the core aspects of combinatorics involves understanding permutations and combinations, both of which describe ways to count arrangements or selections. Let's break them down:Permutations:

• Use permutations when the order matters. Think of arranging books on a shelf where each book's position is crucial, or deciding race placements.
• The formula for permutations of selecting \( r \) items from \( n \) is \( P(n, r) = \frac{n!}{(n-r)!}\), indicating order-sensitive arrangements.

Combinations:

• For combinations, order does not matter. This is typical in scenarios like choosing members for a team or deciding ingredients for a recipe.
• The combination formula, \( C(n, r) = \frac{n!}{r!(n-r)!} \), is crucial when counting without considering order."

In the original exercise, combinations are relevant because we're selecting a committee where the order of choice has no impact. Unlike permutations, combinations only account for distinct groups, not how they are ordered within those groups.