Problem 80 - A Graphical Approach to Precalculus with Limits

Question

Prove each statement for positive integers \(n\) and \(r\), with \(r \leq n\). (Hint: Use the definitions of permutations and combinations.) $$C(n, 1)=n$$

Step-by-Step Solution

Answer

Yes, it is true that \( C(n, 1) = n \).

Step 1: Define the combination formula

The combination formula for selecting \( r \) items from \( n \) items is given by \( C(n, r) = \frac{n!}{r!(n-r)!} \). This is useful for finding the number of combinations of \( r \) items from a total of \( n \) items.

Step 2: Substitute values

Substitute \( r = 1 \) into the combination formula: \( C(n, 1) = \frac{n!}{1!(n-1)!} \).

Step 3: Simplify the equation

Simplify the equation: \( C(n, 1) = \frac{n \cdot (n-1)!}{1! \cdot (n-1)!} \). Cancel the \((n-1)!\) in the numerator and denominator, and simplify to \( C(n, 1) = \frac{n}{1} \).

Step 4: Final result

Since dividing \( n \) by \( 1 \) gives \( n \), we have the final result: \( C(n, 1) = n \).

Key Concepts

Permutations

Permutations are an essential concept in combinatorics that describe the number of ways to arrange a set of items. When you have a group of objects and you want to know how many different ways you can order them, you're looking for permutations. For example, if you have three letters A, B, and C, the permutations are ABC, ACB, BAC, BCA, CAB, and CBA. That's six different ways.
Here's an important point: order matters in permutations. This means that ABC is considered different from CBA, even though they contain the same letters.
To calculate the number of permutations of a set, we use the permutation formula: \[ P(n, r) = \frac{n!}{(n-r)!} \]

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

The exclamation mark (!) denotes the factorial operation, which we'll cover next. The permutation formula allows us to determine the number of different ways we can select and arrange r items from n total items.

Combinations

Combinations are similar to permutations, but with one notable difference: order doesn't matter in combinations. This counts how many ways you can choose a subset from a larger set, where the order of selection doesn't matter.Take a simple example: selecting two letters from A, B, and C. The combinations are AB, AC, and BC. Notice that BA and AB are considered the same combination because, in combinations, the sequence doesn't count.
The formula for finding combinations is:\[ C(n, r) = \frac{n!}{r!(n-r)!} \]

• n is the total number of items.
• r is the number of items to be chosen.

This formula shows how easy it is to determine the number of ways you can choose r items from n possibilities without considering different orders. Understanding this helps in many real-world situations, like forming teams or groups from a larger pool of individuals. The problem we solved earlier, \( C(n, 1) = n \), highlights that selecting one item from n items results in n ways, as seen through the simplified combination formula.

Factorials

Factorials provide a foundation for both permutations and combinations. The factorial of a number is simply the product of all positive integers up to that number. For example, the factorial of 3, denoted as 3!, is calculated as 3 × 2 × 1 = 6.
Factorials grow rapidly with larger numbers, giving a sense of how many ways we can arrange a set of items. Here's how you calculate a factorial:

• 5! = 5 × 4 × 3 × 2 × 1 = 120
• 4! = 4 × 3 × 2 × 1 = 24
• 0! is defined as 1

The versatility of factorials lies in their ability to simplify complex combinatorial calculations, making them easier to manage and understand. By understanding factorials, you can solve permutation and combination problems efficiently, as these concepts heavily rely on the factorial operation for their calculations.