Problem 6 - 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 intercept

Step-by-Step Solution

Answer

It's a permutation with 90,720 possibilities.

Step 1: Determine the Type

The situation involves arranging the letters in the word 'intercept'. Since the order of the letters matters in arrangements, this is a permutation problem.

Step 2: Count the Letters

First, count the total number of letters in the word 'intercept', which is 9 letters.

Step 3: Identify Repeated Letters

In 'intercept', the letter 't' appears twice and the letter 'e' appears twice. We must account for these repetitions in our calculations.

Step 4: Apply Permutation Formula with Repetition

The formula to calculate permutations with repetition is given by: \[ \frac{n!}{p1! \times p2!} \] where \( n \) is the total number of items, and \( p1, p2, \ldots \) are the frequencies of repeated items. Here, \( n = 9 \), \( p1 = 2 \) for 't', and \( p2 = 2 \) for 'e'. Calculate: \[ \frac{9!}{2! \times 2!} = \frac{362880}{4} = 90720 \] This results in 90,720 different permutations.

Key Concepts

Arrangement of Letters

When we talk about arranging letters, it's all about how we can position them in every possible order. Each unique order in which we can rearrange the letters is called a permutation. Think of permutations as different combinations made by shuffling the letters. For instance, the word "cat" can be arranged in various patterns such as "tac," "atc," and "cta." The order in which we arrange these letters changes the pattern completely.

The importance of the exact arrangement is crucial here because even a small change in order results in a completely new pattern. Therefore, when considering the arrangement of letters, always pay attention to how individual elements are sequenced. If the sequence matters, the scenario is indeed a permutation. This principle applies to words, names, or any set of unique items you might need to arrange.

Permutations with Repetition

Sometimes, certain letters in a word repeat. When this happens, we need a special approach to counting permutations known as permutations with repetition. These permutations account for repeated letters, preventing any overestimation of unique arrangements.

To properly calculate permutations with repetition, the formula \[ \frac{n!}{p1! \times p2! \times \ldots} \]
is used. Here, \( n \) represents the total number of letters, while \( p1, p2, \ldots \) are the factorials of the numbers of times each repeating letter appears.

By dividing by the factorials of the repeating letters, we effectively discount duplicate arrangements caused by repetition. This method ensures that only unique permutations are counted, giving you an accurate result. In the case of the word "intercept", since 't' appears twice and 'e' appears twice, these repetitions must be factored into the permutation formula to avoid counting duplicate sequences.

Factorial Calculation

A factorial, symbolized by an exclamation point (\(!\)), is a mathematical operation that multiplies a series of descending natural numbers. For example, \(5!\) means \(5 \times 4 \times 3 \times 2 \times 1\). This operation is crucial for determining permutations, as it calculates the total number of ways to arrange a set number of items.

When dealing with permutations, especially with repeated elements, you'll encounter factorial calculations frequently. In the exercise, we needed to find \(9!\) to start calculating the permutations of the word 'intercept'. Factorials help us comprehend how rapidly the number of arrangements grows as items increase.

The trick with factorials in permutations with repetition is knowing when to divide by other factorials (for each repeated letter). This division prevents overcounting sequences that look different but are actually the same due to repeated letters' presence, ensuring we accurately evaluate permutations.