Calculating the Total Number of Words

Oct 13, 2025 by orovajewels.com 91 views

Introduction

In the realm of combinatorics, we often encounter problems that involve selecting and arranging items from a given set. The question at hand is a classic example of this, where we are tasked with determining the number of words that can be formed using only 5 letters of the word CAMBRIDGE, with the added constraint that all the vowels must be present. This problem requires a deep understanding of combinatorial principles and the application of mathematical formulas to arrive at the solution.

Understanding the Problem

The word CAMBRIDGE has 9 letters, and we are only interested in using 5 of them to form different words. The constraint is that all the vowels must be present in each word. The vowels in the word CAMBRIDGE are A, I, and E. We need to find the number of words that can be formed using these 5 letters, with the condition that all 3 vowels must be included.

Breaking Down the Problem

To approach this problem, we need to break it down into smaller, manageable parts. Let's first consider the selection of the 5 letters from the word CAMBRIDGE. We can use the combination formula to calculate the number of ways to select 5 letters from a set of 9 letters. The combination formula is given by:

nCr = n! / (r!(n-r)!)

where n is the total number of items, r is the number of items to be selected, and ! denotes the factorial function.

In this case, we have n = 9 (total number of letters) and r = 5 (number of letters to be selected). Plugging these values into the combination formula, we get:

9C5 = 9! / (5!(9-5)!) = 9! / (5!4!) = (9 × 8 × 7 × 6 × 5!) / (5! × 4!) = (9 × 8 × 7 × 6) / (4 × 3 × 2 × 1) = 126

So, there are 126 ways to select 5 letters from the word CAMBRIDGE.

Selecting the Vowels

Now that we have selected 5 letters, we need to ensure that all 3 vowels (A, I, and E) are present in each word. We can use the combination formula again to calculate the number of ways to select 3 vowels from the set of 3 vowels. The number of ways to select 3 vowels from 3 vowels is given by:

3C3 = 3! / (3!(3-3)!) = 3! / (3!0!) = 1

So, there is only 1 way to select all 3 vowels.

Selecting the Consonants

We have already selected 3 vowels, and we need to select 2 consonants from the remaining 2 consonants (C, B, R, and D, and G). We can use the combination formula to calculate the number of ways to select 2 consonants from 4 consonants. The number of ways to select 2 consonants from 4 consonants is given by:

4C2 = 4! / (2!(4-2)!) = 4! / (2!2!) = (4 × 3 × 2!) / (2! × 2!) = (4 × 3) / (2 × 1) = 6

So, there are 6 ways to select 2 consonants from the remaining 4 consonants.

Calculating the Total Number of Words

We have selected 3 vowels and 2 consonants, and we need to calculate the total number of words that can be formed using these 5 letters. We can use the multiplication principle to calculate the total number of words. The multiplication principle states that if we have n ways to perform one task and m ways to perform another task, then the total number of ways to perform both tasks is given by n × m.

In this case, we have 1 way to select all 3 vowels and 6 ways to select 2 consonants. Therefore, the total number of words that can be formed using these 5 letters is given by:

1 × 6 = 6

However, this is not the final answer. We need to consider the order of the vowels. When we say there are 6C2 combinations, we assume that the order of the vowels does not matter. But in this case, the order of the vowels does matter. We need to consider the number of ways to arrange the 3 vowels in each word.

Arranging the Vowels

We have 3 vowels (A, I, and E), and we need to arrange them in each word. We can use the permutation formula to calculate the number of ways to arrange the 3 vowels. The permutation formula is given by:

nPr = n! / (n-r)!

where n is the total number of items and r is the number of items to be arranged.

In this case, we have n = 3 (total number of vowels) and r = 3 (number of vowels to be arranged). Plugging these values into the permutation formula, we get:

3P3 = 3! / (3-3)! = 3! / 0! = 3! / 1 = 3 × 2 × 1 = 6

So, there are 6 ways to arrange the 3 vowels in each word.

Calculating the Final Answer

We have selected 3 vowels and 2 consonants, and we have arranged the 3 vowels in each word. We need to calculate the final answer, which is the total number of words that can be formed using these 5 letters. We can use the multiplication principle to calculate the final answer. The final answer is given by:

6 × 6 = 36

Therefore, there are 36 words that can be formed using only 5 letters of the word CAMBRIDGE, with the condition that all 3 vowels must be included.

Conclusion

Q: What is the main difference between this problem and the classic combination problem?

A: The main difference between this problem and the classic combination problem is the constraint that all 3 vowels must be included in each word. In the classic combination problem, the order of the items does not matter, but in this problem, the order of the vowels does matter.

Q: How do we calculate the number of ways to select 3 vowels from 3 vowels?

A: We use the combination formula to calculate the number of ways to select 3 vowels from 3 vowels. The combination formula is given by:

nCr = n! / (r!(n-r)!)

where n is the total number of items and r is the number of items to be selected. In this case, we have n = 3 (total number of vowels) and r = 3 (number of vowels to be selected). Plugging these values into the combination formula, we get:

3C3 = 3! / (3!(3-3)!) = 3! / (3!0!) = 1

So, there is only 1 way to select all 3 vowels.

Q: How do we calculate the number of ways to select 2 consonants from 4 consonants?

A: We use the combination formula to calculate the number of ways to select 2 consonants from 4 consonants. The combination formula is given by:

nCr = n! / (r!(n-r)!)

where n is the total number of items and r is the number of items to be selected. In this case, we have n = 4 (total number of consonants) and r = 2 (number of consonants to be selected). Plugging these values into the combination formula, we get:

4C2 = 4! / (2!(4-2)!) = 4! / (2!2!) = (4 × 3 × 2!) / (2! × 2!) = (4 × 3) / (2 × 1) = 6

So, there are 6 ways to select 2 consonants from the remaining 4 consonants.

Q: How do we calculate the total number of words that can be formed using these 5 letters?

A: We use the multiplication principle to calculate the total number of words that can be formed using these 5 letters. The multiplication principle states that if we have n ways to perform one task and m ways to perform another task, then the total number of ways to perform both tasks is given by n × m.

In this case, we have 1 way to select all 3 vowels and 6 ways to select 2 consonants. Therefore, the total number of words that can be formed using these 5 letters is given by:

1 × 6 = 6

However, this is not the final answer. We need to consider the order of the vowels.

Q: How do we calculate the number of ways to arrange the 3 vowels in each word?

A: We use the permutation formula to calculate the number of ways to arrange the 3 vowels in each word. The permutation formula is given by:

nPr = n! / (n-r)!

where n is the total number of items and r is the number of items to be arranged. In this case, we have n = 3 (total number of vowels) and r = 3 (number of vowels to be arranged). Plugging these values into the permutation formula, we get:

3P3 = 3! / (3-3)! = 3! / 0! = 3! / 1 = 3 × 2 × 1 = 6

So, there are 6 ways to arrange the 3 vowels in each word.

Q: What is the final answer to the problem?

A: The final answer to the problem is 36. This is the total number of words that can be formed using only 5 letters of the word CAMBRIDGE, with the condition that all 3 vowels must be included.

Q: What is the significance of this problem in combinatorics?

A: This problem is significant in combinatorics because it illustrates the application of combination and permutation formulas to solve a real-world problem. It also highlights the importance of considering the order of items when calculating the number of ways to arrange them.

Q: Can this problem be generalized to other words?

A: Yes, this problem can be generalized to other words. The same principles and formulas can be applied to any word that has a specific number of vowels and consonants. The only difference will be the number of vowels and consonants in the word.

Q: What are some real-world applications of combinatorics?

A: Combinatorics has many real-world applications, including:

Computer science: Combinatorics is used in computer science to solve problems related to algorithms, data structures, and software engineering.
Statistics: Combinatorics is used in statistics to analyze and interpret data.
Biology: Combinatorics is used in biology to study the structure and function of biological systems.
Economics: Combinatorics is used in economics to model and analyze economic systems.

These are just a few examples of the many real-world applications of combinatorics.