Problem 4 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Evaluate the given expression. $$ \frac{6 !}{4 ! 2 !} $$

Step-by-Step Solution

Answer

The given expression is \(\frac{6!}{4!2!}\). Evaluating factorials, we have 6! = 720, 4! = 24, and 2! = 2. Plugging these values into the expression, we get \(\frac{720}{(24)(2)} = \frac{720}{48} = 15\). Therefore, the expression evaluates to 15.

Step 1: Understand Factorials

A factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n. For example, 5! means 5 × 4 × 3 × 2 × 1.

Step 2: Evaluate 6!

To find 6!, multiply all the positive integers up to 6: 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

Step 3: Evaluate 4!

To find 4!, multiply all the positive integers up to 4: 4! = 4 × 3 × 2 × 1 = 24

Step 4: Evaluate 2!

To find 2!, multiply all the positive integers up to 2: 2! = 2 × 1 = 2

Step 5: Divide the factorials

Now that we have the values for 6!, 4!, and 2!, we can plug them into our expression and divide: \(\frac{6!}{4!2!} = \frac{720}{(24)(2)}\)

Step 6: Simplify the expression

Now, let's simplify the expression by dividing 720 by 24 and 2: \(\frac{720}{(24)(2)} = \frac{720}{48} = 15\) So, the expression \(\frac{6!}{4!2!}\) evaluates to 15.

Key Concepts

Factorial Notation

In mathematics, the factorial of a non-negative integer is a fundamental concept, often represented with the exclamation mark (!). The factorial function is defined such that for any non-negative integer \( n \) the factorial \( n! \) is the product of all positive integers from 1 to \( n \) itself. This means \( n! = n \times (n-1) \times ... \times 2 \times 1 \). For instance, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

There are special cases to consider, where \( 0! \) is defined to be 1, despite the fact \( 0 \) has no positive integers less than or equal to it. Factorial notation is not just a mathematical curiosity; it comes into play in various areas such as counting permutations and combinations, describing series expansions, and more. Understanding how to evaluate factorials is the first step in solving many problems in combinatorial mathematics.

When evaluating expressions involving factorials, like the given exercise \( \frac{6!}{4!2!} \), it's often easy to overlook the fact that many terms in the numerator and denominator can cancel out. We could also compute \( 6! \) as \( 6 \times 5 \times 4 \times 3 \times 2 \times 1 \), but when we divide by \( 4! \) (which is \( 4 \times 3 \times 2 \times 1 \)) and \( 2! \) (which is \( 2 \times 1 \)), the \( 4 \times 3 \times 2 \times 1 \) of the \( 4! \) and the \( 2 \times 1 \) of the \( 2! \) both fully cancel out those same terms in the \( 6! \), simplifying the calculation significantly.

Evaluating Expressions

Evaluating expressions with factorial notation requires careful manipulation of terms, especially when dealing with division, as demonstrated in the solution to the exercise \( \frac{6!}{4!2!} \). Rather than immediately calculating the factorials and then dividing, which can be inefficient and error-prone, it's advisable to first identify and cancel out any common factors.

In our exercise, after breaking down each factorial into its component products, we notice that \( 6! \) includes the product of all numbers from 1 through 6, but \( 4! \) and \( 2! \) contain some of these numbers as well. This allows us to cancel out common terms, simplifying \( \frac{6!}{4!2!} \) to \( \frac{6 \times 5}{2} \) and ultimately to 15 after performing the arithmetic. Key to evaluating such expressions is using this simplification wherever possible, which not only makes the math easier but also greatly reduces the likelihood of errors in calculation.

Permutation and Combination

Permutations and combinations are concepts in mathematics that deal with the counting of arrangements and selections, respectively. In the language of factorials, permutations of \( n \) objects taken \( r \) at a time are denoted as \( \frac{n!}{(n-r)!} \), and they count the different ways you can order \( r \) items from a set of \( n \) options.

Combinations, on the other hand, are represented as \( \frac{n!}{r!(n-r)!} \), and they tell us how many ways we can select \( r \) items from a set of \( n \) without considering the order. The exercise \( \frac{6!}{4!2!} \) we solved is in fact a combination expression, representing the number of different ways to select 2 items from a collection of 6, without caring about the order of selection.

To contextualize, imagine choosing 2 fruit from a basket of 6 different types; the formula lets us calculate how many distinct pairs we can form. Fundamental to understanding permutations and combinations is recognizing this connection with factorials, as it allows an elegant expression of some otherwise complex counting problems.