Problem 45 - A Graphical Approach to Precalculus with Limits

Question

Use the fundamental principle of counting or permutations to solve each problem. A business school offers courses in keyboarding, spreadsheets, transcription, business English, technical writing, and accounting. In how many ways can a student arrange a schedule if 3 courses are taken?

Step-by-Step Solution

Answer

There are 120 ways to arrange a schedule with 3 courses.

Step 1: Identify Problem Type

To solve this problem, notice that we need to find how many ways 3 courses can be arranged from a total of 6 different courses. This is a permutation problem because the order in which the courses are taken matters.

Step 2: Apply Permutation Formula

The permutation formula for arranging 'r' items from a total of 'n' items is given by \( P(n, r) = \frac{n!}{(n-r)!} \). Here, \( n = 6 \) and \( r = 3 \).

Step 3: Calculate Factorials

Calculate \( 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \) and \( (6-3)! = 3! = 3 \times 2 \times 1 = 6 \).

Step 4: Compute Permutation

Substitute the factorial values into the permutation formula: \( P(6, 3) = \frac{720}{6} \).

Step 5: Simplify Result

Calculate \( \frac{720}{6} = 120 \). Therefore, there are 120 different ways to arrange a schedule taking 3 courses.

Key Concepts

Fundamental Principle of Counting

The fundamental principle of counting is an essential concept in combinatorics. It helps determine the number of ways different events can occur. This principle can be very helpful when solving problems involving permutations or combinations.

It states that if an event occurs in a certain number of ways, say event A in "m" ways, and is followed by another event B that can occur in "n" ways independently of A, then the total number of ways both events can occur is the product of "m" and "n". In a scenario with more events, you would keep multiplying the number of ways each event can occur.

• Event A: can happen in m ways.
• Event B: can occur in n ways.
• Total ways: m × n ways.

To find how students can arrange their schedules by choosing 3 courses out of a total of 6, we have to consider that choosing a course is dependent on the previous choice — making it a permutation problem, not just a multiplication of independent events.

Permutation Formula

A permutation is an arrangement of objects where the order is important. The permutation formula is used to determine the number of possible arrangements of "r" items from a total of "n" items.

The mathematical formula is given by:\[P(n, r) = \frac{n!}{(n-r)!}\]

• "n" is the total number of items.
• "r" is the number of items to arrange.
• "!" (factorial) denotes the product of all positive integers up to that number.

For our exercise, with 6 different courses and needing to arrange 3, we calculate the permutation as follows: use \(n = 6\) and \(r = 3\).

With permutation, unlike combination, every arrangement where the order differs counts as distinct. That's why in our exercise with courses, each unique sequence matters.

Factorial Calculation

Factorials are a crucial part of permutations and combinations, often used in the formula for arranging items. The factorial of a number "n" is denoted as \(n!\) and represents the product of all positive integers up to "n".

Let's calculate some factorials to see how they work.
For instance, for \(6!\), compute as follows:\[6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\]

And for \(3!\), calculate:\[3! = 3 \times 2 \times 1 = 6\]

Factorials grow very quickly as the number increases. You need to calculate them precisely, especially since they often appear in the denominator of the permutation formula.

In our solution, using \(6!\) and \((6-3)! = 3!\), we determine the permutations by substituting these factorials into our permutation formula, from which we get a total of 120 different course arrangements.