Problem 58 - Contemporary Precalculus

Question

(a) Verify that \(\left(\begin{array}{l}7 \\\ 2\end{array}\right)=\left(\begin{array}{l}7 \\ 5\end{array}\right)\) (b) Let \(r\) and \(n\) be integers with \(0 \leq r \leq n .\) Prove that \(\left(\begin{array}{l}n \\ r\end{array}\right)=\left(\begin{array}{c}n \\\ n-r\end{array}\right) .[\text {Note: Part }(\mathrm{a}) \text { is the case when } n=7\) and \(r=2.1\)

Step-by-Step Solution

Answer

Question: Verify that \(\left(\begin{array}{l}7 \\\ 2\end{array}\right)=\left(\begin{array}{l}7 \\\ 5\end{array}\right)\) and prove that \(\left(\begin{array}{l}n \\\ r\end{array}\right)=\left(\begin{array}{c}n \\\ n-r\end{array}\right)\) for integers \(0 \leq r \leq n\). Answer: Verification: We find that \(\left(\begin{array}{l}7 \\\ 2\end{array}\right) = 21\) and \(\left(\begin{array}{l}7 \\\ 5\end{array}\right) = 21\), so they are equal. Proof: We showed that \(\frac{n!}{r!(n-r)!} = \frac{n!}{(n-r)!r!}\) for any integers \(0 \leq r \leq n\), proving that \(\left(\begin{array}{l}n \\\ r\end{array}\right)=\left(\begin{array}{c}n \\\ n-r\end{array}\right)\).

Step 1: Part (a): Verification

To verify that \(\left(\begin{array}{l}7 \\\ 2\end{array}\right)=\left(\begin{array}{l}7 \\\ 5\end{array}\right)\), we will compute the binomial coefficients for each using the formula \(\left(\begin{array}{l}n \\\ r\end{array}\right) = \frac{n!}{r!(n-r)!}\). For \(\left(\begin{array}{l}7 \\\ 2\end{array}\right)\): Using the binomial coefficient formula with \(n=7\) and \(r=2\): \(\left(\begin{array}{l}7 \\\ 2\end{array}\right) = \frac{7!}{2!(7-2)!} = \frac{7!}{2!5!} = \frac{7\times6\times5\times4\times3\times2\times1}{(2\times1)(5\times4\times3\times2\times1)} = \frac{7\times6}{2} = 21\). Now for \(\left(\begin{array}{l}7 \\\ 5\end{array}\right)\): Using the binomial coefficient formula with \(n=7\) and \(r=5\): \(\left(\begin{array}{l}7 \\\ 5\end{array}\right) = \frac{7!}{5!(7-5)!} = \frac{7!}{5!2!} = \frac{7\times6\times5\times4\times3\times2\times1}{(5\times4\times3\times2\times1)(2\times1)} = \frac{7\times6}{2} = 21\). Since \(\left(\begin{array}{l}7 \\\ 2\end{array}\right) = 21\) and \(\left(\begin{array}{l}7 \\\ 5\end{_array}\right) = 21\), we can conclude that \(\left(\begin{array}{l}7 \\\ 2\end{array}\right)=\left(\begin{array}{l}7 \\\ 5\end{array}\right)\).

Step 2: Part (b): Proof

Now, we will prove that \(\left(\begin{array}{l}n \\\ r\end{array}\right)=\left(\begin{array}{c}n \\\ n-r\end{array}\right)\). Consider the binomial coefficients for each: For \(\left(\begin{array}{l}n \\\ r\end{array}\right)\), using the formula \(\left(\begin{array}{l}n \\\ r\end{array}\right) = \frac{n!}{r!(n-r)!}\): For \(\left(\begin{array}{c}n \\\ n-r\end{array}\right)\), using the formula \(\left(\begin{array}{c}n \\\ n-r\end{array}\right) = \frac{n!}{(n-r)!r!}\). From the above expressions for both coefficients, we can see the denominators are the same, with \(r!\) and \((n-r)!\). In fact, the expressions are identical: \(\frac{n!}{r!(n-r)!} = \frac{n!}{(n-r)!r!}\) Hence, we have proved that \(\left(\begin{array}{l}n \\\ r\end{array}\right)=\left(\begin{array}{c}n \\\ n-r\end{array}\right)\).

Key Concepts

Combinatorics

Combinatorics is a fascinating area of mathematics focused on counting, arranging, and analyzing various elements within sets. It's essentially the mathematics of counting. When you're asked how many ways you can pick a certain number of items from a larger collection, you're dealing with a problem rooted in combinatorics.
In combinatorics, we develop formulas and methods to solve complex counting problems without needing to manually list each possibility. For example, if you're deciding how often you can choose 2 pieces of candy out of a jar of 7, combinatorics helps you figure this out easily.
Some key concepts include permutations and combinations, which help calculate the number of possible arrangements and selections. Understanding these concepts can be quite helpful in various fields, like computer science for algorithm efficiency, biology for genetic arrangements, and more.

• **Permutations** refer to arrangements where order matters.
• **Combinations** refer to selections where order doesn't matter.

Understanding these principles lays a strong foundation for tackling more challenging mathematical problems.

Factorials

Factorials, often denoted by the exclamation mark (!), are a fundamental concept in mathematics, especially in combinatorics. A factorial of a non-negative integer \(n\), written as \(n!\), is the product of all positive integers less than or equal to \(n\).
This concept simplifies the process of calculating permutations and combinations.
For instance, \(5!\) means \(5 \times 4 \times 3 \times 2 \times 1 = 120\). The factorial function grows rapidly with larger numbers, making it effective for representing large products succinctly.
Factorials play a critical role in the calculation of binomial coefficients, which we use in problems involving combinations. In a simple scenario, the number of ways to arrange \(n\) items is given by \(n!\). For combinatorial purposes, we often see expressions like \(n!/(r!(n-r)!)\), which are essential for computing how many ways we can select \(r\) items from \(n\) items. This formula is the backbone of many problems dealing with selection and arrangement.

Permutation and Combination

Permutations and combinations are two essential methods used in combinatorics to count the number of ways to arrange or select items from a set.
**Permutations** involve arrangement where order does matter. For example, in how many ways can you arrange 3 books on a shelf? The answer is given by: \(3! = 6\).
**Combinations**, on the other hand, do not take order into account. They answer questions like how many different teams can be made from a group. For instance, choosing 3 friends out of 7 to go on a trip, here order doesn't matter, and we use combinations to solve this.
Mathematically, permutations are calculated using \(n!/(n-r)!\) while combinations use \(n!/(r!(n-r)!)\). It's crucial to decide whether order impacts the count—this choice dictates whether to use permutations or combinations.

• **Example**: Choosing 2 out of 7 letters to form initials is a combination, whereas determining all orderings of 7 letters is a permutation problem.
• **Key Distinction**: Use permutations when the order is important and combinations when it's not.

This distinction helps us apply the correct method and makes problem-solving much clearer and efficient in combinatorial problems.