Problem 78 - 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.) $$C(n, n)=1$$

Step-by-Step Solution

Answer

The expression simplifies to 1, so \(C(n, n) = 1\).

Step 1: Recall the Definition of Combination

The number of combinations of selecting \(r\) objects from \(n\) objects, denoted as \(C(n, r)\), is given by the formula:\[ C(n, r) = \frac{n!}{r!(n-r)!} \] where \(!\) denotes the factorial of a number.

Step 2: Apply the Given Values

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

Step 3: Simplify the Expression

Simplify \((n-n)!\), which is equal to \(0!\). Recall that \(0! = 1\).Thus, the expression for \(C(n, n)\) becomes:\[ C(n, n) = \frac{n!}{n! \cdot 1} = \frac{n!}{n!} \]

Step 4: Calculate the Result

Since any non-zero number divided by itself is \(1\), we have:\[ C(n, n) = 1 \]

Key Concepts

Factorials

Factorials are fundamental in the study of permutations and combinations. A factorial, denoted by the symbol (!), refers to the product of all positive integers up to a given number. The factorial of a non-negative integer **Definition**:- For any non-negative integer \( n \), the factorial of \( n \) is written as \( n! \) and is calculated as: \[ n! = n \times (n-1) \times (n-2) \times \ldots \times 1 \] - By convention, \( 0! = 1 \).**Examples**:- \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)- \( 3! = 3 \times 2 \times 1 = 6 \)In combinatorics, factorials are crucial because they appear in formulas involving counting arrangements and combinations, such as the permutations \( P(n, r) \) and combinations \( C(n, r) \). The factorial indicates all possible ways to arrange or combine a set number of elements from a larger group.

Permutations

Permutations focus on the arrangement of objects in a specific order. The order is important in permutations, differing from combinations where the order doesn't matter.**Definition**:- A permutation of a set is any rearrangement of its elements.- The number of permutations of \( n \) objects taken \( r \) at a time is given by: \[ P(n, r) = \frac{n!}{(n-r)!} \]**Key Points**:- Used when you are interested in the order of outcomes.- For example, calculating possible lineups or rankings.**Examples**:- To arrange 3 books out of 5 on a shelf, calculate \( P(5, 3) \): \[ P(5, 3) = \frac{5!}{(5-3)!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 60 \]Permutations are essential in distinguishing different possible sequences from a group of elements, showing how arrangement affects outcomes.

Combinatorics

Combinatorics is a broader area of mathematics involving counting, arrangement, and organization of objects. It's often broken down into several principal areas, including permutations and combinations.**Combination**:- Combinations involve selecting items from a group, where the order does not matter.- The formula for combinations of \( n \) objects taken \( r \) at a time is: \[ C(n, r) = \frac{n!}{r!(n-r)!} \]**Core Concepts**:- **Choice vs. Arrangement:** Combinations count selections without concern for order, unlike permutations, which are all about arrangement.- **Example:** Choosing 2 fruits from a basket of 3 (apple, orange, banana) results in combinations like (apple, orange), (apple, banana), and (orange, banana). For this, use \( C(3, 2) \) which results in: \[ C(3, 2) = \frac{3!}{2!(3-2)!} = \frac{3 \times 2 \times 1}{2 \times 1 \times 1} = 3 \]In combinatorics, understanding the subtleties between combinations and permutations allows for solving diverse counting problems, from selecting teams to arranging people in a queue. By leveraging the power of factorials, combinatorics offers a structured way to dissect complex arrangements.