Problem 11 - College Algebra with Modeling and Visualization

Question

Calculate the number of distinguishable strings that can be formed with the given number of a's and b's. Four \(a^{\prime}\) 's, four \(b\) 's

Step-by-Step Solution

Answer

There are 70 distinguishable strings of 4 'a's and 4 'b's.

Step 1: Determine Total Arrangements

First, we need to determine the total number of arrangements of 8 letters (4 'a's and 4 'b's). Since we are arranging 8 letters, the total number of arrangements is given by the formula for permutations of a multiset: \[ \frac{n!}{n_1! \times n_2!} \]where \(n\) is the total number of items to arrange, and \(n_1\) and \(n_2\) are the counts of each repeated item, 'a' and 'b' respectively. Here, \(n = 8\), \(n_1 = 4\), and \(n_2 = 4\).

Step 2: Apply the Permutation Formula

Substitute the values into the permutation formula:\[ \frac{8!}{4! \times 4!} \]Calculate the factorials: \[ 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320 \] \[ 4! = 4 \times 3 \times 2 \times 1 = 24 \]

Step 3: Perform the Division

Now, substitute the calculated factorial values into the formula:\[ \frac{40320}{24 \times 24} \]Calculate \(24 \times 24 = 576\).

Step 4: Final Calculation

Finish the calculation by dividing 40320 by 576:\[ \frac{40320}{576} = 70 \]Thus, the number of distinguishable strings that can be formed is 70.

Key Concepts

Factorials

Factorials are an essential mathematical concept used in combinatorics. A factorial, denoted by an exclamation mark (!), is a product of all positive integers up to a given number. For example, for any positive integer \( n \), the factorial of \( n \) is represented as \( n! \).

• For \( n = 3 \), \( 3! = 3 \times 2 \times 1 = 6 \).
• \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

Factorials are widely used in permutations and combinations to determine the number of possible arrangements. They grow very rapidly with larger numbers; for example, \( 10! = 3,628,800 \). In this context, factorials help calculate the total ways to arrange a set of items, such as letters in a string.

Permutations

Permutations are ways to arrange a set of items. When you are asked to calculate permutations, you are looking to number the different ways to order a given set of items.When there are no repeated elements, you typically use the formula \( n! \), where \( n \) is the number of items. However, the presence of repeated elements, as in multiset permutations, alters this.In a multiset permutation, you're dealing with groups of indistinguishable items. We use the formula:\[\frac{n!}{n_1! \times n_2! \times \, ... \, \times n_k!}\]where \( n_1, n_2, ..., n_k \) represent the quantities of each repeated item type.For example, in arranging the string with 4 'a's and 4 'b's, the permutations are calculated as \(\frac{8!}{4! \times 4!}\), accounting for repeats of each letter. Thus, permutation strikes a balance in counting only the distinct arrangements.

Multisets

A multiset, unlike a set, allows for the repetition of elements. In combinatorics, multisets are particularly interesting because they complicate the counting of permutations due to repeated elements.The classic problem of determining unique string arrangements from repeated letters, like 4 'a's and 4 'b's, exemplifies the utility of multisets. Regular sets wouldn't be able to account for these repetitions effectively.When dealing with multisets, each type of item needs careful consideration. A handy formula involves dividing the total factorial arrangements \( n! \) by the product of factorials of repeated items. For example, to find the number of unique sequences of letters in a multiset with 8 total letters (4 'a's and 4 'b's), you would use:\[\frac{8!}{4! \times 4!}\]Understanding multisets allows us to accurately count unique permutations of elements that include repetitions, essential in problems with repeated elements like our string arrangement example.