Problem 55 - Algebra 2 and Trigonometry

Question

The Electronic Depot received 108 \(\mathrm{mp} 3\) players, 12 of which are defective. In how many different ways can 6 players be selected for display such that: a. none are defective? b. half are defective? c. all are defective?

Step-by-Step Solution

Answer

a. 676,133,800 ways b. 31,433,600 ways c. 924 ways

Step 1: Calculate total non-defective players

The total number of non-defective players is found by subtracting the number of defective players from the total number of players: 108 total players - 12 defective players = 96 non-defective players.

Step 2: Calculate combinations for none defective

To find the number of ways to choose 6 non-defective players out of 96, we use the combination formula: \[ \binom{96}{6} \]Calculating this, we get:\[ \binom{96}{6} = \frac{96 \times 95 \times 94 \times 93 \times 92 \times 91}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 676,133,800 \] So, there are 676,133,800 ways.

Step 3: Calculate combinations for half defective

We need to select 3 defective players and 3 non-defective players. First, calculate the combinations for each:- Defective: \( \binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220 \)- Non-defective: \( \binom{96}{3} = \frac{96 \times 95 \times 94}{3 \times 2 \times 1} = 142,880 \)Now, multiply the two results to find the total number of ways to select 3 defective and 3 non-defective players:\( 220 \times 142,880 = 31,433,600 \).So, there are 31,433,600 ways.

Step 4: Calculate combinations for all defective

To select all 6 players as defective, we first note that there are only 12 defective players in total. The number of ways to choose all 6 from these is:\[ \binom{12}{6} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 924 \]So, there are 924 ways.

Key Concepts

Combination Formula

In the context of combinatorics, the combination formula is essential when you need to determine the number of ways to choose a subset of items from a larger set, where the order of selection does not matter. It's distinct from permutations, where order does matter. The combination formula is expressed as follows:

1. \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \)

Here:

• \( n \) is the total number of items.
• \( r \) is the number of selected items.
• \( n! \) (n factorial) is the product of all positive integers up to \( n \).

Using this formula, you can solve various selection problems, such as the one posed by the exercise, where different scenarios of selecting defective and non-defective items are evaluated.

Defective Products

In many practical applications, distinguishing between defective and non-defective items is crucial, especially in inventory control and quality assurance. In this exercise, the concept of defective products is critical as it represents items that do not meet standard specifications. When selecting items, the occurrence and number of defective products can alter the total combinations and outcomes significantly. For example:

• When none of the selected items is defective, only the non-defective items are used for combinations.
• When half of the items are defective, combinations must include both defective and non-defective products.
• When all selected items are defective, only combinations of the defective items are considered.

Understanding defective products helps in developing strategies to minimize defects and optimize selection processes.

Binomial Coefficient

The binomial coefficient, denoted as \( \binom{n}{r} \), is a fundamental component of the combination formula. It's called a 'binomial' because it emerges in the binomial theorem, which describes the algebraic expansion of powers of a binomial expression.In combinatorics:

• The binomial coefficient quantifies the number of ways to choose \( r \) elements from a set of \( n \) elements without regard to order.
• It encapsulates the idea of "choosing subsets" and is widely used for probability calculations and decision-making scenarios.

In our exercise, the binomial coefficient is directly used to compute the various selection scenarios, offering insights into real-life decision making.

Selection Problem

In combinatorial problems, a selection problem requires determining how many ways you can choose a subset of items from a larger set, while considering certain conditions or constraints. In our case, the selection problem is illustrated through choosing MP3 players for display under different conditions of defectiveness. The problem is broken down into three main scenarios:

• Selecting 6 players where none are defective, involves choosing entirely from the pool of non-defective players.
• Selecting 3 defective and 3 non-defective players requires a mixed approach and emphasizes balance.
• Selecting all 6 players as defective checks for the least possible combinations available.

Understanding these selection constraints and their implications helps in strategizing solutions in both theoretical problems and practical inventory management.