Problem 57 - Contemporary Precalculus

Question

(a) Verify that \(\left(\begin{array}{l}9 \\ 1\end{array}\right)=9\) and \(\left(\begin{array}{l}9 \\ 8\end{array}\right)=9\) (b) Prove that for each positive integer \(n,\left(\begin{array}{l}n \\\ 1\end{array}\right)=n\) and \(\left(\begin{array}{c}n \\\ n-1\end{array}\right)=n .[\text {Note: Part }(\mathrm{a}) \text { is the case when } n=9\) and \(n-1=8.1\)

Step-by-Step Solution

Answer

Question: Verify the result \(\binom{n}{1} = n\) and \(\binom{n}{n-1} = n\) for the specific case \(n=9\), and prove that this holds for all positive integers \(n\).

Step 1: (Calculate the Binomial Coefficient for \({9\choose 1}\) and \({9\choose 8}\))

First, calculate the values of \({9\choose 1}\) and \({9\choose 8}\). The binomial coefficient is given by the formula: \({n\choose k}=\frac{n!}{k!(n-k)!}\). For \({9\choose 1}\): \({9\choose 1}=\frac{9!}{1!(9-1)!}=\frac{9!}{1!8!}=\frac{9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{(1)(8\times 7\times 6\times 5\times 4\times 3\times 2\times 1)}=9\) For \({9\choose 8}\): \({9\choose 8}=\frac{9!}{8!(9-8)!}=\frac{9!}{8!1!}=\frac{9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{(8\times 7\times 6\times 5\times 4\times 3\times 2\times 1)(1)}=9\) Hence, for \(n=9\), \({n\choose 1} = n\) and \({n\choose n-1} = n\), as \(\left(\begin{array}{l}9 \\\ 1\end{array}\right)=9\) and \(\left(\begin{array}{l}9 \\\ 8\end{array}\right)=9\). ##Step 2: General Proof##

Step 2: (Prove for General Positive Integer \(n\))

Now, we need to prove that \({n\choose 1} = n\) and \({n\choose n-1} = n\) for every positive integer \(n\). To prove \({n\choose 1} = n\), consider: \({n\choose 1}=\frac{n!}{1!(n-1)!}=\frac{n\times(n-1)\times(n-2)\times\cdots\times 2\times 1}{(1)((n-1)\times(n-2)\times \cdots\times 2\times 1)}=n\) Hence, \({n\choose 1} = n\) for every positive integer \(n\). To prove \({n\choose n-1} = n\), consider: \({n\choose n-1}=\frac{n!}{(n-1)!(1)!}=\frac{n\times(n-1)\times(n-2)\times\cdots\times 2\times 1}{((n-1)\times(n-2)\times \cdots\times 2\times 1)(1)}=n\) Hence, \({n\choose n-1} = n\) for every positive integer \(n\). In conclusion, we have proven that for every positive integer \(n\), \({n\choose 1} = n\) and \({n\choose n-1} = n\).

Key Concepts

Combinatorics

Combinatorics is a branch of mathematics that deals with counting and arranging objects. It often involves finding the number of ways to choose or arrange items in a specific order or combination. In the context of the exercise, we're dealing with the concept of binomial coefficients, denoted as \({n \choose k}\). This notation represents the number of ways to choose \(k\) items from a total of \(n\) items without regard to order.
Combinatorics uses various tools and formulas to calculate these values, and one of the most common formulas is the binomial coefficient formula. This formula is particularly useful for solving problems related to probability and statistics. The deeper understanding of combinatorics not only enhances problem-solving skills but also lays the foundation for advanced mathematical studies.

• Binomial coefficient: \({n \choose k} = \frac{n!}{k!(n-k)!}\)
• Components: total objects \(n\) and chosen objects \(k\)

Factorials

Factorials play a crucial role in combinatorics and are used to calculate permutations and combinations, including binomial coefficients. The factorial of a positive integer \(n\), denoted \(n!\), is the product of all positive integers from 1 to \(n\). It is defined as:
\[ n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1 \]
For example, the factorial of 5 is calculated as follows:
5! = 5 × 4 × 3 × 2 × 1 = 120.
Factorials are instrumental in simplifying the fraction in the binomial coefficient formula. By dividing the factorial of \(n\) by the product of the factorials of \(k\) and \(n-k\), we get the number of ways to choose \(k\) items from \(n\) items.
Understanding the concept of factorials is essential as it helps in learning combinatorial identities and simplifies the process of solving complex mathematical problems.

• Notation: \(n!\)
• Applications: permutations, combinations

Positive Integers

Positive integers are the set of natural numbers starting from 1, extending infinitely. They are represented by the symbol \( \mathbb{Z}^+ \) and include numbers such as 1, 2, 3, and so on. In combinatorics and the binomial coefficient, positive integers serve as the basis for calculations.
In the given exercise, positive integers \(n\) and various factorial calculations depend on these numbers to determine the number of combinations. Understanding the use of positive integers is significant in problem-solving within mathematics and computer science, as they form the basis for various algorithms and counting techniques.
Positive integers help in defining elements in sets, calculating permutations, and determining the number of ways to arrange items or perform selections.

• Examples: 1, 2, 3, ...
• Use: defining set elements, calculations in algorithms

Mathematical Proof

Mathematical proof is a logical argument that demonstrates the truth of a given statement based on previously established facts and axioms. In mathematics, proofs are indispensable for validating formulas and concepts.
In our problem, we used mathematical proof to validate the relation between the binomial coefficients and positive integers, showing that \({n \choose 1} = n\) and \({n \choose n-1} = n\) for any positive integer \(n\).
The proof involves breaking down the formula into understandable parts and verifying each step using algebraic manipulation and simplification. This approach ensures that the solution is comprehensive, rigorous, and universally applicable.
Having a good grasp of mathematical proofs helps in developing logical reasoning and problem-solving abilities, which are critical skills in mathematics and related fields.

• Key Components: logic, axioms, previously established facts
• Importance: ensures validity, enhances understanding