Problem 46 - Algebra and Trigonometry

Question

Show that \(\left(\begin{array}{l}{n} \\ {0}\end{array}\right)=1\) and \(\left(\begin{array}{l}{n} \\ {n}\end{array}\right)=1\)

Step-by-Step Solution

Answer

Both expressions equal 1 as each selects no objects or all objects from \( n \).

Step 1: Understanding Binomial Coefficients

The binomial coefficient is written as \( \left( \begin{array}{c} n \ k \end{array} \right) \) and it represents the number of ways to choose \( k \) objects from a set of \( n \) objects without regard to the order of selection. It is also known by the formula: \( \left( \begin{array}{c} n \ k \end{array} \right) = \frac{n!}{k!(n-k)!} \)

Step 2: Evaluate \( \left( \begin{array}{c} n \\ 0 \end{array} \right) \)

For the case when \( k = 0 \), the formula for the binomial coefficient becomes \( \left( \begin{array}{c} n \ 0 \end{array} \right) = \frac{n!}{0!(n-0)!} = \frac{n!}{n!} = 1 \). Hence there is exactly one way to choose no objects from \( n \).

Step 3: Evaluate \( \left( \begin{array}{c} n \\ n \end{array} \right) \)

For the case when \( k = n \), the formula for the binomial coefficient becomes \( \left( \begin{array}{c} n \ n \end{array} \right) = \frac{n!}{n!(n-n)!} = \frac{n!}{n!} = 1 \). This illustrates that there is exactly one way to choose all \( n \) objects from \( n \).

Key Concepts

Combinatorics

Combinatorics is the field of mathematics focused on counting, arranging, and grouping objects in sets. It plays a significant role when it comes to selecting subsets from a larger set, and this is where binomial coefficients come into play. With combinations, we can ask questions like, "How many ways are there to select a team of 3 from 10 players?" or "How many different groups of 5 can be formed from a set of 20 items?" Combinatorics gives us the tools to solve such problems efficiently.

In the context of our original exercise, combinatorics helps us understand why the binomial coefficients \( \left(\begin{array}{c}{n} \ {0}\end{array}\right) \) and \( \left(\begin{array}{c}{n} \ {n}\end{array}\right) \) are both equal to 1. The logic is simple: there is only one way to choose no objects from a set (which is to choose nothing), and only one way to choose all objects (taking everything). These scenarios illustrate basic principles of combinatorics, emphasizing the understanding of "choosing" and "arranging."

Factorials

Factorials are a mathematical operation denoted by an exclamation point (!). The factorial of a non-negative integer \( n \) is the product of all positive integers less than or equal to \( n \). In simpler terms, \( n! \) can be understood as

• If \( n = 0 \), then \( 0! = 1 \) by definition.
• The factorial of any other number \( n \), such as \( n = 5 \), would be \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

In the exercise, the use of factorials helps in calculating binomial coefficients. When calculating \( \left( \begin{array}{c} n \ 0 \end{array} \right) \), we utilize the fact that \( 0! = 1 \). This simplifies our formula: \( \left( \begin{array}{c} n \ 0 \end{array} \right) = \frac{n!}{0!(n-0)!} = \frac{n!}{n!} = 1 \). This simple definition helps ensure calculations work smoothly and accurately when it comes to combinatorial problems.

Binomial Theorem

The Binomial Theorem is a powerful tool in combinatorics, allowing for the expansion of powers of sums. It tells us how to expand a binomial expression \((a + b)^n\) and involves the use of binomial coefficients. In the formula for expanding \((a + b)^n\), the coefficients of the expansion terms \(a^k b^{n-k}\) are given by binomial coefficients:\[(a + b)^n = \sum_{k=0}^{n} \left(\begin{array}{c} n \ k \end{array}\right) a^{n-k} b^k\]
In our exercise, we explored specific values of binomial coefficients, notably \( \left(\begin{array}{c} n \ 0 \end{array}\right) \) and \( \left(\begin{array}{c} n \ n \end{array}\right) \), where both are shown to be 1. This demonstrates the way coefficients work in expansions. For example, \((x + 1)^n\) includes the term \(x^n\) and the constant term 1, both with a coefficient of 1, representing the cases examined in the exercise. Understanding these grounds us in the fundamental applications of the theorem for polynomial expansions.