Question

The binomial coefficient \(\left(\begin{array}{l}6 \\ 4\end{array}\right)\) equals (a) \(\frac{6 !}{4 !}\) (b) \(\frac{6 !}{4 ! \cdot 2 !}\) (c) \(\frac{6 !}{2 !}\) (d) \(\frac{(6-4) !}{2 !}\)

Short answer

Option (b) \( \frac{6!}{4! \times 2!} \)

Step-by-step solution

Understand the Binomial Coefficient

The binomial coefficient \(\binom{n}{k}\) is calculated using the formula \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\). Let's determine what values of \(n\) and \(k\) are in this context.

Identify Values

Here, \(n = 6\) and \(k = 4\). Plug these values into the formula.

Substitute Values into Formula

Using the formula \(\binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6!}{4! \times 2!}\). This is the correct form to calculate the given binomial coefficient.

Match with Given Options

Compare this result with the provided options to find the match: (a) \(\frac{6!}{4!}\) (b) \(\frac{6!}{4! \times 2!}\) (c) \(\frac{6!}{2!}\) (d) \(\frac{(6-4)!}{2!}\). Option (b) is exactly the same as our result.

Key concepts

Binomial Theorem

The Binomial Theorem is a formula that explains how to expand expressions raised to a power, like \( (a + b)^n\). When expanded, it produces a series of terms that include binomial coefficients. These coefficients are represented as \( \binom{n}{k} \). The theorem itself can be written as:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
Here, each term \( \binom{n}{k} \) represents the number of ways to choose k elements from n elements without regard to order. This process involves factorials and combinatorics, which will be explained further.
The Binomial Theorem has various uses in combinatorial mathematics, probability, and algebra, simplifying complex polynomial expansions.

Factorials

A factorial, denoted with an exclamation point (e.g., \( n! \)), is a mathematical operation where you multiply a sequence of descending positive integers. For example:

• \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
• \( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
• \( 0! = 1 \) (by definition)

Factorials are integral to determining binomial coefficients. For example, in the problem \( \binom{6}{4} \), the formula \( \frac{6!}{4!(6-4)!} \) uses factorials to find the result. Identifying the factorials needed and performing the arithmetic correctly is key to solving such problems.

Combinatorics

Combinatorics is a branch of mathematics dealing with counting, arrangement, and combination of objects. In the context of the given problem, combinatorics helps us understand how to calculate the number of ways to choose a subset of items from a larger set.
For example, \( \binom{6}{4} \) asks the question: How many ways can you choose 4 items out of 6? Using combinatorics principles, we determine this number with combinations.
Combinatorial calculations rely on the binomial coefficient formula:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
Here, \( n \) is the total number of items, and \( k \) is the number of items to choose. This formula helps ensure that each selection scenario is counted exactly once.

Mathematical Problem Solving

Problem-solving is a multi-step process in mathematics. To solve problems involving the binomial coefficient, follow these general steps:

• Identify the problem and understand what is being asked. For instance, in the exercise, you need to find \( \binom{6}{4} \).
• Recognize key values and terms. Here, \( n = 6 \) and \( k = 4 \).
• Apply relevant formulas or equations, such as substituting into \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \).
• Perform the necessary calculations, simplifying where possible to get the correct result.
• Compare your result with provided options or interpret the result contextually.

Mastery of such problem-solving techniques can greatly enhance efficiency and accuracy in mathematics.