Problem 5 - College Algebra

Question

In Exercises 1-8, evaluate the given binomial coefficient. $$\left(\begin{array}{l}6 \\ 6\end{array}\right)$$

Step-by-Step Solution

Answer

The value of the binomial coefficient \(\left(\begin{array}{l}6 \ 6\end{array}\right)\) is 1.

Step 1: Understanding the binomial coefficient

The binomial coefficient, often read as 'n choose k', defines the number of ways to choose \(k\) elements from a set of \(n\) elements. It's generally defined as \(\left(\begin{array}{l} n \ k \end{array}\right) = \frac{n!}{k!(n-k)!}\), where '!' denotes factorial. Factorial represents the product of all positive integers up to that number. For example, \(4! = 4*3*2*1 = 24\). When \(n = k\), \(\left(\begin{array}{l} n \ k \end{array}\right) = 1\).

Step 2: Evaluate the binomial coefficient

In the given problem, we've \(\left(\begin{array}{l} 6 \ 6 \end{array}\right)\). According to the above principle, since \(n = k\), this binomial coefficient equals 1.

Key Concepts

Factorial

The concept of a factorial is foundational in various branches of mathematics, including combinatorics and probability theory. A factorial, denoted by an exclamation point (!), is the product of all positive integers up to a certain number. To be more precise, for any non-negative integer, \( n \), the factorial is defined as: \[ n! = n \times (n-1) \times (n-2) \times \ldots \times 3 \times 2 \times 1 \. \] For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). One special case that often confuses students is zero factorial, which is defined as \( 0! = 1 \).

Understanding factorials is crucial when dealing with permutations and combinations as they often appear in the formulas that allow us to calculate the number of ways an event can occur. It is also a recurring element in the calculation of binomial coefficients.

Combinatorics

Combinatorics is a vast area of mathematics that deals with counting, both as a means and an end in understanding the structures of various types. It includes studying sequences, graph theory, and binomial coefficients. Combinatorial problems ask us to figure out the number of ways certain patterns can be formed.

One fundamental principle in combinatorics is the rule of sum and rule of product. The rule of sum states that if there are \( a \) ways to do something and \( b \) ways to do another thing, and these two actions cannot be done at the same time, then there are \( a + b \) ways to choose one of these actions. The rule of product states that if there are \( a \) ways to do something and \( b \) ways to do another thing after the first action is done, then there are \( a \times b \) ways to perform both actions in sequence.

These principles are immensely useful for solving problems involving binomial coefficients where we count the number of combinations of items.

Permutations and Combinations

Permutations and combinations are two concepts that deal with the arrangements of elements within a set. Despite their close relationship, they deal with different scenarios.

• Permutations refer to arrangements of items where the order matters. For instance, the arrangement of letters in a word. The formula for calculating permutations of \( n \) items taken \( k \) at a time is given by \( P(n, k) = \frac{n!}{(n-k)!} \).


• Combinations, on the other hand, refer to arrangements where the order does not matter. They are calculated using the binomial coefficient, which is the topic of our exercise. The formula for the number of combinations of \( n \) items taken \( k \) at a time is exactly the binomial coefficient \( \left(\begin{array}{c} n \ k \end{array}\right) = \frac{n!}{k!(n-k)!} \).

Understanding the difference between permutations and combinations is fundamental in correctly analyzing scenarios that involve grouping or arranging objects, which is a frequent task in probability and counting problems.