Problem 31 - Calculus for Biology and Medicine

Question

Five A's, three \(B\) 's, and six \(C\) 's are to be arranged into a 14 letter word. How many different words can you form?

Step-by-Step Solution

Answer

168,168 different words can be formed.

Step 1: Understanding the Problem

We have 14 letters in total, consisting of five A's, three B's, and six C's. We need to calculate the number of distinct arrangements of these letters.

Step 2: Using the Permutation Formula for Repeated Items

When we have repeated items in a permutation problem, we use the formula for permutations of multiset: \[ \frac{n!}{n_1! \times n_2! \times ... \times n_k!} \]where \( n \) is the total number of items, and \( n_1, n_2, ..., n_k \) are the frequencies of the repeated items. For this problem, \( n = 14 \), \( n_1 = 5 \) (A's), \( n_2 = 3 \) (B's), and \( n_3 = 6 \) (C's).

Step 3: Applying the Formula

Substitute the values into the formula:\[ \frac{14!}{5! \times 3! \times 6!} \]We need to calculate this fraction to find the number of different arrangements.

Step 4: Calculating 14!

Calculate \( 14! \):\[ 14! = 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 87,178,291,200 \]

Step 5: Calculating the Denominator

Calculate \( 5! \), \( 3! \), and \( 6! \):\[ 5! = 120, \quad 3! = 6, \quad 6! = 720 \]Multiply these:\[ 5! \times 3! \times 6! = 120 \times 6 \times 720 = 518,400 \]

Step 6: Final Calculation

Now, divide the results:\[ \frac{87,178,291,200}{518,400} = 168,168 \]This is the number of distinct arrangements of the letters.

Key Concepts

Combinatorics

Combinatorics is a fascinating branch of mathematics that deals with counting arrangements, combinations, and groupings. It's all about figuring out how to arrange or combine different elements in a set. The key to combinatorics is understanding how different elements can be grouped or arranged without necessarily repeating any of them.

In our scenario with letters, combinatorics helps us determine how many different ways we can arrange a given set of letters. If you imagine each letter as a distinct entity, combinatorics teaches us to consider factors such as repetition and order. By using permutation and combination formulas, we can solve problems that involve counting the different ways elements can be organized. This is extremely helpful when dealing with complex arrangements, like determining how many permutations of a word can exist when letters are repeated.

Combinatorics isn't just about math puzzles—it has real-world applications in areas like cryptography, data organization, and even biology. Understanding its principles allows for better problem-solving strategies in these various fields.

Factorial Calculation

Factorial calculation is a method in mathematics that takes an integer and multiplies it by every positive integer less than it. It's denoted as the number followed by an exclamation mark, like this: \( n! \).

For example, when we calculate \( 5! \), we take:

• \( 5 \times 4 \times 3 \times 2 \times 1 \)

This results in 120. Factorials grow very quickly, which makes them particularly powerful when calculating possible permutations or combinations in a set.

In our problem, we used \( 14! \) because we had 14 letters to arrange. Each time, the factorial represents the total number of ways to arrange a set of distinct items.

• This calculation involves multiplying all positive integers up to 14.

When dealing with problems that involve repeated elements, factorial calculations help us understand the total number of unique permutations of those elements. Factorials are a fundamental aspect of permutations and combinations, providing a straightforward way to comprehend these concepts.

Multiset Permutation Formula

The multiset permutation formula is an essential tool in combinatorics when dealing with repeated elements. In situations where certain elements appear more than once, like in our exercise with the letters, we must adjust our approach to count permutations accurately.

The formula is as follows:

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

Here, \( n \) is the total number of items (such as letters), and \( n_1, n_2, \ldots, n_k \) represent the frequencies of each distinct element that repeats within the set. So, these terms break down the total set to account for duplicate items.

In our problem, using five A's, three B's, and six C's meant that we couldn't just shuffle 14 distinct letters. Instead, we needed the multiset permutation formula to reduce the count for duplicates, giving us the exact number of unique permutations. It helps us see through the clutter of repetitive elements and reach a clear understanding of how many distinct arrangements are possible. Understanding and applying this formula is crucial when calculating permutations in any scenario involving repeated items.