Problem 18 - College Algebra with Corequisite Support

Question

For the following exercises, compute the value of the expression. $$ P(9,6) $$

Step-by-Step Solution

Answer

The value of the expression \( P(9,6) \) is 60480.

Step 1: Understanding Permutations

The expression \( P(n, r) \) represents a permutation, which is the number of ways to arrange \( r \) objects from a set of \( n \) objects. The formula to compute a permutation is given by \( P(n, r) = \frac{n!}{(n-r)!} \).

Step 2: Identify Values for n and r

From the given expression \( P(9,6) \), identify \( n = 9 \) and \( r = 6 \). These values will be used in the permutation formula.

Step 3: Calculate Factorials in the Formula

Calculate \( n! = 9! \) and \( (n-r)! = 3! \) (since \( (n-r) = 9-6 = 3 \)).- \( 9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362880 \)- \( 3! = 3 \times 2 \times 1 = 6 \)

Step 4: Apply the Permutation Formula

Substitute the calculated factorial values into the permutation formula:\[ P(9,6) = \frac{9!}{3!} = \frac{362880}{6} \].

Step 5: Compute the Final Result

Divide the factorial of 9 by the factorial of 3:\[ \frac{362880}{6} = 60480 \]. Therefore, \( P(9,6) = 60480 \).

Key Concepts

Factorials

A factorial, denoted by an exclamation point (e.g., 9!), is a mathematical operation that multiplies a number by all the positive integers less than itself. Factorials are key in permutations and combinations because they help calculate the total number of ways objects can be arranged.
To calculate the factorial of a number:

• Start with the number itself.
• Multiply it by every decreasing whole number until you reach 1.

For example, 9! equals 9 multiplied by 8, 7, and so on, down to 1. This is written as:
\[9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362880\]
Similarly, 3! is 3 multiplied by 2 and 1, calculated as:
\[3! = 3 \times 2 \times 1 = 6\]
Factorials grow very fast, which makes them perfect for problems involving large sets of data. They turn abstract arrangements into solid mathematics.

Combinatorics

Combinatorics is a field of mathematics focused on counting, both as a means and an end in obtaining results, and certain properties of finite structures. It's concerned with understanding how different arrangements of objects in a set can occur.
Permutations are a part of combinatorics. They help us figure out the total number of unique ways to arrange a particular number of objects taken from a larger group. The permutation formula \( P(n, r) \) is used when the order of the objects matters. Here, \( n \) represents the total set of objects, and \( r \) represents the number we want to select and arrange.
The permutation formula is:
\[P(n, r) = \frac{n!}{(n-r)!}\]
This formula divides the full arrangement (given by \( n! \)) by the arrangement of the remaining objects \((n-r)!\). In our example, \( P(9, 6) \) meant choosing 6 objects from a set of 9 and arranging them. This is why permutations are handy in problems where the sequence is essential, such as racing results or seating arrangements.

Algebraic Expressions

Algebraic expressions are mathematical phrases that can include numbers, variables, and operations. They are used to represent real-world problems in a mathematical form. In the context of permutations, algebraic operations help derive formulas that express relationships between numbers and operations.
An expression like \( P(9, 6) \) in permutations includes variables and factorial symbols to represent complex arrangements simply. Here's how you break it down:

• \( n \) and \( r \) are variables (in our case, \( n = 9 \) and \( r = 6 \)).
• \( n! \) and \((n-r)!\) are factorials used as part of the formula.

The permutation expression \( \frac{9!}{3!} \) becomes a simple algebraic form that can be solved step by step. First, we calculate the factorials, then divide them as seen in the original solution, aiding a systematic approach to what could otherwise be a daunting calculation.
Understanding how to manipulate algebraic expressions, like in this permutation example, allows us to efficiently solve and simplify a variety of mathematical problems.