Problem 44 - Precalculus with Limits

Question

In Exercises 43 - 46, find the number of distinguishable permutations of the group of letters. \( B, B, B, T, T, T, T, T, \)

Step-by-Step Solution

Answer

The number of distinguishable permutations of the given group of letters is calculated to be \( \frac{8!}{3! \times 5!} \).

Step 1 Counting the Total Number of Letters

Firstly, count the number of letters in total. We have 3 B's and 5 T's. So, the total number of letters is \(n = 3 + 5 = 8\)

Step 2 Determine the Number of Same Letters

Next, determine the number of each type of letter. We have 3 B's and 5 T's. So, the multiple numbers are \(n1 = 3\) and \(n2 = 5\) respectively.

Step 3 Apply the Formula for Multiset Permutations

Apply the formula for permutations of a multiset, which is \[ \frac{n!}{n1! \times n2!} \].

Step 4 Perform the Calculations

To solve for the number of distinguishable letter permutations, perform the calculations. After calculating factorial for all, we have: \[ \frac{8!}{3! \times 5!} \] After calculating, you'll find the answer.

Key Concepts

Multiset Permutations

A fundamental concept in combinatorics is the idea of multiset permutations. When we talk about permutations, we usually refer to distinct items being arranged in different ways. But what happens when some items are identical? That's where multiset permutations come into play.

A multiset is a collection where each item can appear more than once. When we want to find how many different ways we can arrange these items, we use the formula for multiset permutations. The formula is important because it accounts for repetitions of the same item. For a multiset with total items represented by \( n \), and subsets of varying identical items being \( n_1, n_2, \) etc., the formula for permutations becomes:

• \( \frac{n!}{n_1! \times n_2! \times \ldots} \).

This formula effectively "divides out" the redundancy caused by identical items, providing the number of unique arrangements.

Factorial Calculation

Factorials play a crucial role in permutation calculations, especially multiset permutations. Mathematically, a factorial is the product of all positive integers up to a specified number. It's denoted by an exclamation mark. For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

In the context of permutations, we use factorials to calculate the total possible arrangements of a set of items. The formula for permutations of a multiset relies on factorial calculations to account for both the total number of items and the number of identical items.

• Total arrangements: \( n! \)
• Dividing by identicals: \( n_1! \), \( n_2! \), etc.

This allows us to differentiate between truly unique and repeated arrangements in the multiset, offering clear insights into the number of distinguishable permutations.

Combinatorics

Combinatorics is the broader mathematical field that studies counting, arrangement, and combination principles, which are essential in solving permutation problems like our original exercise. It's especially useful in understanding how to group or arrange objects, which can be particularly complex with repeated elements.

In permutation problems involving multisets, combinatoric principles guide us through the process of distinguishing between arrangements by using specific formulas and techniques. This includes:

• Understanding when and how to apply permutation formulas.
• Recognizing the impact of repeating elements in calculations.
• Integrating factorial concepts to streamline difficult counting problems.

Through combinatorics, we gain a framework for dealing with these challenging counting tasks, making sense of how identical components affect overall arrangements in a given set or problem.