Problem 47 - Precalculus: Mathematical for Calculus

Question

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

Step-by-Step Solution

Answer

\(\binom{n}{1} = \binom{n}{n-1} = n\). Both expressions simplify to \( n \).

Step 1: Understanding Binomial Coefficient Notation

The expression \( \binom{n}{k} \) is the binomial coefficient, which represents the number of ways to choose \( k \) objects from \( n \) objects. It is also commonly referred to as "n choose k."

Step 2: Formula for Binomial Coefficients

The formula for the binomial coefficient is \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \( n! \) (n factorial) is the product of all positive integers up to \( n \).

Step 3: Evaluating \( \binom{n}{1} \)

Using the formula, we calculate \( \binom{n}{1} = \frac{n!}{1!(n-1)!} = \frac{n \cdot (n-1)!}{1 \cdot (n-1)!} = n \). This simplifies to \( n \) because the \((n-1)!\) terms cancel out.

Step 4: Evaluating \( \binom{n}{n-1} \)

Similarly, \( \binom{n}{n-1} = \frac{n!}{(n-1)!1!} = \frac{n \cdot (n-1)!}{(n-1)! \cdot 1} = n \). Again, the \((n-1)!\) terms cancel out, leaving us with \( n \).

Step 5: Concluding Both Equalities

We have shown that both \( \binom{n}{1} \) and \( \binom{n}{n-1} \) simplify to \( n \). Therefore, \( \binom{n}{1} = \binom{n}{n-1} = n \).

Key Concepts

Combinatorics

Combinatorics is an area of mathematics that deals with counting, arranging, and finding patterns in sets of objects. When you think about problems such as deciding how many different ways you can choose a certain number of items from a larger set, you're entering the realm of combinatorics. This field of math helps us understand complex arrangements and selections by breaking them down into more manageable formulas and principles.

One of the key techniques in combinatorics is using the binomial coefficient, often described in problems where you need to choose a subset of objects from a larger set. Situations where this is common include organizing teams from groups or creating different playlist combinations from a music library. Through combinatorics, such seemingly challenging questions become a series of simpler calculations, allowing more straightforward solutions and a deeper understanding of the problem structure.

Factorial

The concept of a factorial is crucial when dealing with binomial coefficients and many combinatorics problems. A factorial, denoted by the exclamation point symbol (!) after a number, represents the product of all positive integers up to that number. For example, the factorial of 5, written as 5!, is calculated as follows:

\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]

Factorials are useful because they can help manage and simplify the calculations of combinations and permutations by providing a way to express arrangements in a more compact form. When working with binomial coefficients, factorials allow us to compute things like "5 choose 2" efficiently by using the formula for combinations. Understanding factorials provides the foundation needed to delve deeper into other combinatorial concepts, ensuring you can find precise and correct results easily.

Combinations

Combinations are a fundamental part of combinatorics, involving the selection of items from a larger set where the order does not matter. The number of ways this selection can happen is what we calculate using combinations. This is captured in mathematical form by the binomial coefficient \( \binom{n}{k} \).

For instance, if you wanted to choose 2 fruits from a set of 3 different fruits (say, apples, bananas, and cherries), the combinations would consider which two fruits you choose, but not the order they are picked. In such cases, we calculate the essential combinations by using:

\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]

Here, the crucial elements involve understanding when and how factorials factor into the simplification of the results. This creates a bridge to how real-world selections work, like forming teams from a group of players without considering the position of each player on the list. Recognizing how these combinations work directly leads to solutions that are logical and practical.