Problem 14 - Algebra and Trigonometry

Question

Find the number of distinguishable permutations of the given letters. $$ A B C D D D E E $$

Step-by-Step Solution

Answer

The number of distinguishable permutations is 3360.

Step 1: Determine Total Number of Letters

Count the total number of letters in the sequence. We have 8 letters: \( A, B, C, D, D, D, E, E \).

Step 2: Identify Repeating Letters

Recognize the repeating letters and their frequencies: D appears 3 times, and E appears 2 times.

Step 3: Apply Permutation Formula for Distinguishable Permutations

The formula for the number of distinguishable permutations of a multiset is given by:\[\frac{n!}{n_1! \cdot n_2! \cdot \ldots \cdot n_k!}\]where \( n \) is the total number of letters and \( n_1, n_2, \ldots, n_k \) are the frequencies of repeating letters.In this case:\[ n = 8, \, n_D = 3, \, n_E = 2 \]

Step 4: Calculate Factorials for Repeating Letters

Calculate the factorial of the repeating groups:\[ 3! = 6 \] for D's, and\[ 2! = 2 \] for E's.

Step 5: Apply the Formula

Substitute the calculated values into the formula:\[\text{Number of permutations} = \frac{8!}{3! \times 2!} \]Simplify it step-by-step:\[8! = 40320\]\[3! \times 2! = 6 \times 2 = 12\]Thus,\[\frac{40320}{12} = 3360\].

Key Concepts

Distinguishable Permutations

In the world of permutations, the term "distinguishable permutations" refers to the arrangement of objects or elements where some are identical. Imagine having a set of letters, such as those in this exercise: \(A, B, C, D, D, D, E, E\). If each letter were unique, calculating permutations would be straightforward; you'd simply arrange all letters in numerous possible ways.

However, when letters (or objects) repeat, they must be accounted for, leading to the need for distinguishable permutations. This scenario means that permutations that seem different due to identical elements must be considered the same. For example, swapping any of the three D's doesn't lead to a new arrangement.

• Think of distinguishable permutations as finding unique arrangements of an object set with repeated elements.
• Distinguishable permutations prevent overcounting identical arrangements.

Using the correct formula ensures accurate results and keeps consideration of the actual unique permutations available.

Multiset Permutation Formula

To identify the number of distinguishable permutations in a collection where objects repeat, we apply the multiset permutation formula. This formula calculates the total number of ways to order a set while accounting for duplicates. The formula is:

\[ \frac{n!}{n_1! \, n_2! \, \ldots \, n_k!} \]

Where:

• \(n\) is the total number of items in the set.
• \(n_1, n_2, \ldots, n_k\) are the frequencies of the repeated items.

For example, in the word sequence \(A, B, C, D, D, D, E, E\), we have:

• Total letters \(n = 8\)
• D repeats \(n_D = 3\) times
• E repeats \(n_E = 2\) times

Plug these values into the formula to find the permutations' count:

\[ \text{Number of permutations} = \frac{8!}{3! \, 2!} \]
This formula helps in preventing errors that arise from treating repeated objects as unique.

Factorials

Factorials play a crucial role in finding permutations, including those of distinguishable and multiset permutations. A factorial of a number \( n \), denoted as \( n! \), is the product of all positive integers less than or equal to \( n \). In simple terms, it's like multiplying a number by each number smaller than it, down to 1.

• For instance, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).

In our exercise with \(A, B, C, D, D, D, E, E\), we needed the factorials of the total letters and the repeats:

• \(8!\) helps us understand the total permutations as if no element repeats.
• \(3!\) for the repeating D's, so \(3! = 3 \times 2 \times 1 = 6\).
• \(2!\) for the repeating E's, so \(2! = 2 \times 1 = 2\).

Applying these factorials helps adjust the count to reflect only unique permutations, by dividing the total permutations by those formed by repeats.

Combinatorics

Combinatorics is the area of mathematics focused on counting, combinations, and permutations. This field allows us to explore all possible arrangement scenarios of a set of objects. When applied to distinguishable permutations, combinatorics provides necessary tools to calculate configurations accurately without repetitive counting errors.

An understanding of combinatorics is crucial in problems where not all elements are unique, as in our exercise where D and E are repeated. The strategies involve:

• Recognizing the total items and the subset of repeating elements.
• Applying combinatorics formulas like the multiset permutation formula to account for repeats.

This exercise demonstrates combinatorics by systematically handling repeated elements while still considering all potential unique configurations, leading to clarity and precision in permutation counting.