Problem 24 - Algebra 2

Question

Determine whether each situation involves a permutation or a combination. Then find the number of possibilities. an arrangement of the letters in the word parallel

Step-by-Step Solution

Answer

Permutation, 3360 arrangements.

Step 1: Determine the Problem Type

First, identify whether the problem is a permutation or a combination. Here, we are dealing with the arrangement of letters, and the order of letters matters. Therefore, this is a permutation problem.

Step 2: Count the Total Number of Letters

The word 'parallel' consists of 8 letters.

Step 3: Identify and Handle Repetitions

In the word 'parallel', some letters repeat. Specifically, there are 3 'l's and 2 'a's.

Step 4: Apply Permutation Formula with Repetitions

For permutation problems with repeating items, use the formula: \[\frac{n!}{p_1! \times p_2! \times \ldots \times p_k!}\]where \(n\) is the total number of items, and \(p_1, p_2, \ldots, p_k\) are frequencies of the repeating items. For 'parallel', it will be:\[\frac{8!}{3! \times 2!}\]

Step 5: Calculate the Factorials and Solve

Calculate the factorials and substitute into the formula:\[8! = 40320, \quad 3! = 6, \quad 2! = 2\]Substitute into the formula:\[\frac{40320}{6 \times 2} = \frac{40320}{12} = 3360\]

Key Concepts

Combinations

Combinations refer to selecting items from a group where the order does not matter. This is different from permutations, where order is crucial. To better understand, imagine you're picking a team of three people from a group of ten. It doesn't matter in which order you select them, what matters is who is ultimately on the team. Generally, the formula for combinations is given by: \[ C(n, k) = \frac{n!}{k!(n-k)!} \] where:

• \( n \) is the total number of items,
• \( k \) is the number of items to select, and
• \( C(n, k) \) represents the number of possible combinations.

Unlike permutations, combinations do not account for different arrangements of the same group of items. For example, the selection \( \{A, B, C\} \) is the same as \( \{C, A, B\} \). Although this exercise is about permutations, understanding combinations helps to draw a clear distinction between these two concepts.

Factorials

Factorials are a crucial part of solving permutations and combinations problems.A factorial, denoted by an exclamation mark, (!), represents the product of all positive integers less than or equal to a given number.For instance, \( n! = n \times (n-1) \times (n-2) \times \, ... \, \times 1 \).Here are some simple examples:

• \( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
• \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
• 0! is defined as 1, which is a unique case and very useful in mathematical calculations.

Factorials are not only used in finding permutations and combinations but also in many other areas of mathematics.The formula for permutations with repetition, for example, makes extensive use of factorials to consider the different possible arrangements.

Repeating Items

When items repeat in a set, it adds an extra layer of complexity to permutation calculations.In a word like 'parallel', these repeating items need special attention."If all items were unique, permutation calculations would be straightforward.Luckily, there's a method to handle this. To find the number of permutations of a set that includes repeating items, we count the total number of items and use the formula: \[ \frac{n!}{p_1! \times p_2! \times \ldots \times p_k!} \] where:

• \( n \) is the total number of items,
• \( p_1, p_2, \ldots, p_k \) are the factorials of the numbers of each repeating item.

In 'parallel', with 3 'l's and 2 'a's, this formula ensures those repeating letters don't artificially inflate the total number of permutations.It simplifies the problem and helps calculate the correct number of unique arrangements.

Permutation Formula

The permutation formula is essential in calculating arrangements where order matters.In contexts where items repeat, the formula must be adjusted to avoid overcounting these duplicates. The standard permutation formula for a set of distinct items is:\[ P(n) = n! \] When repetitions are present, we use:\[ \frac{n!}{p_1! \times p_2! \times \ldots \times p_k!} \]In this version, \( n \) is the total number of items, and \( p_1, p_2, \ldots, p_k \) are the counts of each repeating item.For the word 'parallel', which has repeating letters like 'a' and 'l', the calculations must account for these repetitions using the modified formula. This provides a comprehensive and accurate count of unique permutations, ensuring no possible arrangement is left out while avoiding duplicates. Overall, the permutation formula's power lies in its ability to handle both simple and complex arrangements effectively.