Letter Combinations: Solve Permutations & Combinations Easily

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Hey guys! Let's break down this classic combinatorics problem step-by-step. We're dealing with the letters A, B, C, D, and E, and we want to figure out how many different three-letter combinations we can make. This involves understanding permutations and combinations, which are fundamental concepts in mathematics, especially in probability and statistics. Let's dive into each part of the problem to make sure we grasp every detail. This is going to be a fun journey, so buckle up and let’s get started!

A. How many combinations of three letters can be formed?

To tackle the first part, we need to determine the total number of three-letter combinations that can be formed from the letters A, B, C, D, and E, allowing for repetition. This means we can use the same letter more than once in a combination (e.g., AAA, AAB). When dealing with combinations where repetition is allowed, we use a different approach than when repetition is not allowed. The formula for combinations with repetition is a bit different, and understanding it is key to solving this problem correctly. Let’s break down how to apply this formula in our case.

Understanding Combinations with Repetition

The concept of combinations with repetition might seem a bit tricky at first, but it's actually quite logical once you grasp the underlying principle. Imagine you have n types of items (in our case, 5 letters: A, B, C, D, E) and you want to choose r items (in our case, 3 letters) allowing for repetition. The formula to calculate this is:

C(n + r - 1, r) = (n + r - 1)! / (r! * (n - 1)!)

Where:

  • n is the number of types of items (letters).
  • r is the number of items you want to choose (letters in the combination).
  • '!' denotes the factorial, which means multiplying a number by all positive integers less than it (e.g., 5! = 5 × 4 × 3 × 2 × 1).

This formula may look intimidating, but it’s quite manageable once we plug in our values and work through the calculations step by step. The key is to understand what each variable represents and how they fit into the overall formula. This formula essentially accounts for all the different ways you can select items, considering that you can pick the same item multiple times.

Applying the Formula to Our Problem

In our case, we have:

  • n = 5 (the letters A, B, C, D, and E)
  • r = 3 (we want to form three-letter combinations)

Now, let's substitute these values into the formula:

C(5 + 3 - 1, 3) = C(7, 3)

This simplifies to:

C(7, 3) = 7! / (3! * (7 - 3)!)

Now we need to calculate the factorials and simplify the expression. Let’s break it down further to make sure we don't miss any steps.

Calculating the Factorials

First, let's calculate the factorials:

  • 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
  • 3! = 3 × 2 × 1 = 6
  • (7 - 3)! = 4! = 4 × 3 × 2 × 1 = 24

Now, substitute these values back into the formula:

C(7, 3) = 5040 / (6 * 24)

Next, we perform the multiplication in the denominator:

6 * 24 = 144

So the expression becomes:

C(7, 3) = 5040 / 144

Finally, we perform the division to get our result.

Final Calculation

Now, let's divide 5040 by 144:

5040 / 144 = 35

So, there are 35 combinations of three letters that can be formed from the letters A, B, C, D, and E, allowing for repetition. This means we can create 35 different three-letter combinations, including those with repeated letters. This is a significant number, and it highlights the power of combinations with repetition in expanding the possibilities.

Conclusion for Part A

Therefore, the answer to part A is 35. We found this by using the formula for combinations with repetition and carefully working through the calculations. This result gives us a solid foundation as we move on to the next parts of the problem, where we'll explore combinations with distinct letters and those with at least two repeated letters. Understanding this first step is crucial for grasping the subsequent concepts, so make sure you've got it down before moving on. Next, we'll tackle the combinations without repetition, which will give us a different perspective on the problem.

B. How many combinations of three distinct letters can be formed?

Now, let's dive into part B of our problem: how many combinations of three distinct letters can be formed from the set A, B, C, D, and E? This changes the game a bit because we're no longer allowed to repeat letters within a single combination. For instance, ABC is a valid combination, but AAB is not. This constraint simplifies the calculation in some ways, as we’ll be using the standard combination formula without repetition. However, it's crucial to understand when and how to apply this formula correctly. Let's break it down step by step to make sure we're on the right track.

Understanding Combinations without Repetition

When we talk about combinations without repetition, we're essentially asking: how many ways can we choose a group of items from a larger set, where the order doesn't matter, and we can't pick the same item more than once? This is a classic scenario in combinatorics and has a specific formula to help us calculate the possibilities. The formula for combinations without repetition is:

C(n, r) = n! / (r! * (n - r)!)

Where:

  • n is the total number of items (in our case, 5 letters).
  • r is the number of items we want to choose (in our case, 3 letters).
  • '!' denotes the factorial, as we discussed earlier.

This formula is designed to count the number of ways to choose r items from n items without considering the order and without allowing repetition. It’s a powerful tool for solving a wide range of problems, from card games to selecting teams. Now, let’s apply this formula to our specific problem and see what we get.

Applying the Formula to Our Problem

In this part of the problem, we have:

  • n = 5 (the letters A, B, C, D, and E)
  • r = 3 (we want to form three-letter combinations)

Let’s plug these values into the formula:

C(5, 3) = 5! / (3! * (5 - 3)!)

This formula sets up our calculation, but we still need to compute the factorials and simplify the expression. This is where the step-by-step breakdown becomes crucial. Let’s calculate the factorials first to make the process more manageable.

Calculating the Factorials

Let's calculate the factorials we need:

  • 5! = 5 × 4 × 3 × 2 × 1 = 120
  • 3! = 3 × 2 × 1 = 6
  • (5 - 3)! = 2! = 2 × 1 = 2

Now, we’ll substitute these values back into the formula:

C(5, 3) = 120 / (6 * 2)

Next, let's simplify the denominator by performing the multiplication.

Simplifying the Expression

Now, let’s multiply the numbers in the denominator:

6 * 2 = 12

So our expression becomes:

C(5, 3) = 120 / 12

Finally, we perform the division to get our answer. This will give us the number of combinations of three distinct letters that can be formed from our set.

Final Calculation

Now, let's divide 120 by 12:

120 / 12 = 10

So, there are 10 combinations of three distinct letters that can be formed from the letters A, B, C, D, and E. This result is significantly smaller than the 35 combinations we found in part A, which allowed for repetition. This highlights the impact of the constraint of distinct letters. Now, let's summarize our findings for part B.

Conclusion for Part B

Therefore, the answer to part B is 10. We arrived at this answer by using the formula for combinations without repetition and carefully calculating the result. This part of the problem emphasizes the importance of understanding the constraints in a combinatorial question. Now that we’ve tackled both combinations with and without repetition, let’s move on to the final part of the problem, which asks about combinations with at least two repeated letters. This will tie together what we’ve learned so far and give us a complete picture of the possibilities.

C. Out of the total combinations of three letters, how many have at least two repeated letters?

Finally, we arrive at part C: out of the total combinations of three letters, how many combinations have at least two repeated letters? This question is a clever way to test our understanding of both combinations with and without repetition. We've already calculated the total number of combinations with repetition (35 in part A) and the number of combinations without repetition (10 in part B). To find the number of combinations with at least two repeated letters, we can use a simple subtraction method. This involves taking the total combinations and subtracting the combinations with no repetitions. Let's break down this process step by step.

The Subtraction Method

The key to solving this part of the problem is recognizing that the combinations with at least two repeated letters are essentially what's left over when we remove the combinations with no repetitions from the total combinations. This is a common strategy in combinatorics: sometimes, it's easier to calculate the complement of what you're looking for and subtract it from the total. In this case, the complement is the combinations with no repeated letters. The formula for this approach is:

Combinations with at least two repetitions = Total combinations - Combinations with no repetitions

We already have both of these values from parts A and B, so we can plug them into the formula and get our answer. This method not only simplifies the calculation but also reinforces our understanding of the different types of combinations we've been working with.

Applying the Values from Parts A and B

From part A, we know the total number of three-letter combinations (allowing repetition) is 35. From part B, we know the number of three-letter combinations with distinct letters (no repetition) is 10. Now, let's use these values in our subtraction formula:

Combinations with at least two repetitions = 35 - 10

This is a straightforward subtraction that will give us the number of combinations with at least two repeated letters. Let’s perform this calculation to get our final answer for this part of the problem.

Final Calculation

Performing the subtraction, we get:

35 - 10 = 25

So, there are 25 combinations of three letters that have at least two repeated letters. This means that out of the 35 possible combinations, 25 of them will have either two or three letters that are the same. This result highlights the significant impact of allowing repetition in combinations. Now, let’s wrap up our findings and draw a conclusion for part C.

Conclusion for Part C

Therefore, the answer to part C is 25. We found this by subtracting the number of combinations with no repetitions from the total number of combinations. This approach demonstrates how understanding different types of combinations can help solve more complex problems. By breaking down the problem into smaller parts and using the results we've already calculated, we were able to arrive at a clear and accurate answer.

Final Thoughts and Key Takeaways

Alright guys, we've successfully navigated this combinatorics problem, and it's time to wrap things up. We started with a set of letters and explored the different ways we could form three-letter combinations, considering both cases where repetition was allowed and where it wasn't. We also tackled the specific question of how many combinations have at least two repeated letters. By breaking the problem down into smaller parts, using the appropriate formulas, and understanding the underlying concepts, we were able to find clear answers to each question.

Recap of Our Solutions

Let's quickly recap our solutions:

  • A. Total combinations of three letters (with repetition): 35
  • B. Combinations of three distinct letters (no repetition): 10
  • C. Combinations with at least two repeated letters: 25

These results show the wide range of possibilities that can arise in combinatorics and the importance of carefully considering the constraints of the problem. Whether we're dealing with combinations with repetition, combinations without repetition, or specific conditions like having at least two repeated letters, the key is to understand the underlying principles and apply the appropriate formulas.

The Importance of Understanding Combinations and Permutations

Combinations and permutations are fundamental concepts in mathematics with wide-ranging applications. They're essential in fields like probability, statistics, computer science, and even everyday decision-making. Understanding how to calculate and interpret combinations and permutations can help us make more informed decisions and solve complex problems. By working through problems like this one, we strengthen our understanding of these concepts and become better problem-solvers.

Final Words

So, there you have it! We've explored the world of letter combinations, and hopefully, you've gained a better understanding of how to approach these types of problems. Remember, the key is to break down the problem, identify the constraints, and apply the appropriate formulas. Keep practicing, and you'll become a combinatorics pro in no time! Thanks for joining me on this mathematical adventure, and I'll see you in the next one! Keep those numbers crunching! 😉