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

There are 168,168 different words possible.

Step 1: Understand the Problem

We want to determine how many distinguishable arrangements of letters can be formed using five A's, three B's, and six C's. This is a classic combinatorial permutation problem where we have identical objects in each category.

Step 2: Find the Total Number of Letters

First, calculate the total number of letters in our 'word': There are five A's, three B's, and six C's, which total up to:\[ 5 + 3 + 6 = 14 \] letters.

Step 3: Use the Permutation Formula for Indistinguishable Objects

The formula to find the permutations of a set with indistinguishable objects is:\[\frac{n!}{n_1! imes n_2! imes n_3!}\]Where:- \( n \) is the total number of items (14 in this case),- \( n_1 \) is the number of A's (5),- \( n_2 \) is the number of B's (3),- \( n_3 \) is the number of C's (6).

Step 4: Calculate Factorials Needed

Calculate the factorials needed:\[ n! = 14! \]\[ n_1! = 5! \]\[ n_2! = 3! \]\[ n_3! = 6! \]

Step 5: Compute Factorials

Calculate each factorial:- \( 14! = 87,178,291,200 \)- \( 5! = 120 \)- \( 3! = 6 \)- \( 6! = 720 \)

Step 6: Plug Values into the Formula

Using the permutation formula:\[\frac{14!}{5! imes 3! imes 6!} = \frac{87,178,291,200}{120 imes 6 imes 720}\]

Step 7: Simplify the Expression

Simplify the denominator:- \( 5! \times 3! \times 6! = 120 \times 6 \times 720 = 518,400 \)Now, calculate:\[\frac{87,178,291,200}{518,400} = 168,168\]

Step 8: Interpret the Result

The calculation shows that there are 168,168 different ways to arrange the letters to form a 14-letter 'word' using five A's, three B's, and six C's.

Key Concepts

Factorial Calculation

One of the key concepts in permutation problems involving repetition of items is factorial calculation. Factorials help in determining the number of ways we can arrange a set of items. In simple terms, the factorial of a number, denoted as \( n! \), is the product of all positive integers up to that number.For instance:

• \( 5! \) means multiply all integers from 1 to 5: \( 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
• \( 3! \) is \( 3 \times 2 \times 1 = 6 \)
• \( 6! \) translates to \( 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \)

Factorials grow very quickly as numbers increase. That's why they are crucial in calculations involving large sets, such as in our problem where the total number of letters is 14. Calculating \( 14! \) might seem daunting, but it's manageable with a calculator, resulting in a massive 87,178,291,200 possible arrangements if all were distinguishable.

Combinatorial Permutation

Permutations in combinatorics deal with finding the different ways of arranging a set of items. When dealing with identical items, as in our exercise, we need to engage in combinatorial permutations which account for indistinguishable items through division by factorials.We use the formula:\[\frac{n!}{n_1! \times n_2! \times n_3!}\]Here is why this works:

• \( n! \) calculates all possible arrangements if every item was unique.
• Dividing by \( n_1!, n_2!, \) and \( n_3! \) accounts for repeated items in each category, like all A's, B's, and C's in this set.

This adjustment ensures we count only unique configurations, saving computation from counting the same configuration multiple times. In our case, with 14 letters in total, having five A's, three B's, and six C's leads to the computed total possible permutations being 168,168.

Arrangement of Letters

When tasked with arranging a set of letters into a 'word,' especially with repeated letters, the approach narrows down to correctly applying permutation concepts. To arrange our unique-word pattern using 14 letters, understanding the need for combining repetition with permutation is essential. The initial assumption might be placing each letter as distinct:

• If all 14 letters were different, use \( 14! \).
• However, with repetitions, use the formula to correct over-counts.

Hence, the formula \( \frac{14!}{5! \times 3! \times 6!} \) provides the suitable count. This thoughtful adjustment highlights that the presence of repeated letters doesn't just reduce the number of unique sequences but modifies our approach to counting! Employing this organized method gives us all possible distinct words structured by the provided repetitive letters.