Problem 25 - Calculus for Biology and Medicine

Question

Four A's and five B's are to be arranged into a nine-letter word. How many different words can you form?

Step-by-Step Solution

Answer

There are 126 different words that can be formed.

Step 1: Understanding the Problem

We have a total of nine letters: four 'A's and five 'B's. We need to find out in how many distinct ways we can arrange these letters to form a nine-letter word.

Step 2: Using Combinatorial Formula

The number of ways to arrange a set of letters with repetitions is calculated using the formula: \( \frac{n!}{k_1! \times k_2!} \) where \( n \) is the total number of letters, \( k_1 \) is the number of 'A's, and \( k_2 \) is the number of 'B's.

Step 3: Calculating Factorials

First, calculate the factorials involved:- \( n! = 9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362880 \).- \( k_1! = 4! = 4 \times 3 \times 2 \times 1 = 24 \).- \( k_2! = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

Step 4: Substituting into the Formula

Now, substitute the factorials into the formula: \[ \frac{9!}{4! \times 5!} = \frac{362880}{24 \times 120} = \frac{362880}{2880} \].

Step 5: Performing the Division

Perform the division to find the number of arrangements: \[ \frac{362880}{2880} = 126 \].

Step 6: Conclusion

Hence, there are 126 distinct ways to arrange the letters to form a nine-letter word.

Key Concepts

Factorial

Factorial is a fundamental concept in combinatorics and mathematics. It is represented by the symbol "!" and is the product of all positive integers up to a given number. For example:

• The factorial of 5, denoted as \(5!\), is calculated as \(5 \times 4 \times 3 \times 2 \times 1 = 120\).
• Similarly, \(4! = 4 \times 3 \times 2 \times 1 = 24\).

Factorials play a crucial role in calculating permutations and combinations. They help in determining how many different ways you can arrange or select items from a group. In our case, we used factorials to handle the repeating letters ('A' and 'B') in the word arrangement challenge. By calculating \(9!\), \(4!\), and \(5!\), we found the total number of possible unique arrangements of the letters.

Permutations with Repetition

Permutations with repetition occur when items are arranged but certain items are repeated in the set. This kind of permutation allows for the repeated items to be indistinguishable in the arrangement.To calculate permutations with repetition, especially in cases where you have repeating items, a formula is used:\[\frac{n!}{k_1! \times k_2! \times \ldots \times k_m!}\]where:

• \(n\) is the total number of items.
• \(k_1, k_2, \ldots, k_m\) are the frequencies of the repeating items.

For the given problem, this formula helped us determine how many distinct permutations of the nine letters can exist, given that 'A' and 'B' are repeated four and five times, respectively. Using this formula, we calculated the total permutation by dividing \(9!\) by the factorial of the counts of 'A's and 'B's.

Arrangements of Letters

When arranging letters into words, especially under constraints like repeated letters, we look at how these letters can be ordered to form distinct words. In this specific exercise, each unique arrangement of the nine letters 'A' and 'B' forms a potential "word." The process begins with identifying the total length of the word, which includes all instances of each letter. Given that we had four 'A's and five 'B's, we aimed to discover every potential sequence these could form. This is a classic combinatorial task where repetition constraints must be considered. By applying our permutations with repetition formula, we calculated how many different ways nine letters can be arranged while accounting for their repetitions. This practical application underlines the use of combinatorial formulas in solving real-world word arrangement issues, providing a systematic approach to count distinct configurations.