Problem 71 - Algebra 2

Question

Determine whether each situation involves a permutation or a combination. Then find the number of possibilities. choosing 3 different types of snack foods out of 7 at the store to take on a trip

Step-by-Step Solution

Answer

Combination; 35 ways.

Step 1: Identify the Problem

First, we need to determine whether this is a permutation or a combination problem. In this scenario, we are choosing 3 different types of snack foods from a total of 7 types. Since the order in which we choose the snacks does not matter, we are dealing with a combination problem.

Step 2: Using the Combination Formula

For combination problems, we use the formula: \[C(n, r) = \frac{n!}{r!(n-r)!}\]where \(n\) is the total number of items to choose from (7 types of snacks), and \(r\) is the number of items to choose (3 snacks).

Step 3: Insert Values into the Formula

Insert the values into the combination formula:\[C(7, 3) = \frac{7!}{3!(7-3)!} = \frac{7!}{3! \cdot 4!}\]

Step 4: Calculate the Factorials

Calculate the factorial values:- \(7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040\)- \(3! = 3 \times 2 \times 1 = 6\)- \(4! = 4 \times 3 \times 2 \times 1 = 24\)

Step 5: Complete the Calculation

Using the factorial values calculated in Step 4, compute the combination:\[C(7, 3) = \frac{5040}{6 \cdot 24} = \frac{5040}{144} = 35\]

Key Concepts

Permutations

Permutations involve arranging items in a specific order. It is important to note that order matters in permutations. You can think of permutations like seating arrangements. If you have a list of people and you're assigning them to seats, the order in which you arrange them differs each time, creating different permutations.

• For example, arranging 3 people, A, B, and C, into seats results in different permutations: ABC, ACB, BAC, BCA, CAB, and CBA.

The formula for calculating permutations is given by: \[ P(n, r) = \frac{n!}{(n-r)!} \] where:

• \( n \) is the total number of items.
• \( r \) is the number of items to arrange.

Unlike combinations where the order is irrelevant, permutations require you to pay attention to the sequence of arrangement.

Factorials

Factorials are used heavily in both permutations and combinations. The factorial of a number is the product of all positive integers up to that number. For instance,

• \(4! = 4 \times 3 \times 2 \times 1 = 24 \)
• \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)

The concept of factorial simplifies the calculation of permutations and combinations by breaking down larger computations into manageable parts. It underpins much of the mathematical foundation in counting and arrangement problems.

Combination Formula

Combinations are used when the order of items does not matter. The combination formula allows us to find out how many ways we can select items from a total set. The formal expression for combinations is: \[ C(n, r) = \frac{n!}{r!(n-r)!} \]Here:

• \( n \) represents the total number of items available.
• \( r \) represents the number of items to choose.

Consider the exercise where you are choosing 3 snacks out of 7 available at the store. This is a clear case for using combinations because the sequence in which you pick does not change the outcome. You use the formula \( C(7, 3) = \frac{7!}{3!(7-3)!} = 35 \), which tells us there are 35 different ways to choose the snacks.

Probability

Probability is a mathematical concept used to predict the likelihood of an event occurring. It's a measure expressed on a scale from 0 (impossible event) to 1 (certain event). When dealing with permutations and combinations, probability can provide insight into how likely it is that a particular arrangement or selection occurs.
Probability can be calculated by dividing the number of favorable outcomes by the total number of possible outcomes. For example:

• If you're choosing 3 snacks out of 7, and you want to calculate the probability of picking snacks A, B, and C, you would need the number of favorable outcomes divided by the total combinations, \( \frac{1}{35} \) if each selection is equally likely.

Understanding probability is critical in determining how certain or uncertain a particular scenario might be when making random selections from a defined set.