Problem 27 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Classify each problem according to whether it involves a permutation or a combination. In how many ways can nine different books be arranged on a shelf?

Step-by-Step Solution

Answer

This problem involves a permutation since the arrangement of the books matters. Using the permutation formula \(P(n,r) = n! / (n-r)!\), we find there are \(P(9,9) = 362,880\) ways to arrange nine different books on a shelf.

Step 1: Understand the problem

We need to arrange nine different books on a shelf. Since the order in which the books are arranged matter, this is a permutation problem.

Step 2: Use the permutation formula

The permutation formula is given by: \(P(n,r) = n!/(n-r)!\) where 'n' is the total number of objects to choose from, and 'r' is the number of objects to arrange at a time. In this case, we have 9 different books to arrange (n=9) and we want to arrange all of them (r=9). So the formula becomes: \(P(9,9) = 9!/(9-9)!\)

Step 3: Calculate the factorial values

We need to calculate the values of 9! and (9-9)! in the formula. 9! = 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 362,880 (9-9)! = 0! = 1 (since the factorial of 0 is defined as 1)

Step 4: Solve for the number of ways to arrange the books

Now that we have the factorial values, we can plug them back into the permutation formula: \(P(9,9) = 9! / (9-9)!\) \(P(9,9) = 362,880 / 1\) P(9,9) = 362,880 There are 362,880 ways to arrange nine different books on a shelf.

Key Concepts

Factorial Calculation

The concept of a factorial is a fundamental building block in counting principles and is commonly denoted with an exclamation mark (!). A factorial of a non-negative integer, denoted as \( n! \), represents the product of all positive integers less than or equal to \( n \). For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). Factorials are crucial in solving many mathematical problems, especially in permutations and combinations, as they help determine the number of possible arrangements or selections. The special case of \( 0! \) is defined as \( 1 \) for mathematical consistency and to simplify computations, particularly in permutation and combination formulas.

• For any positive integer \( n \), the factorial \( n! \) is the product of all integers from 1 to \( n \).
• Factorials are used to calculate permutations where the order of objects is important.
• The factorial of zero, \( 0! \), is equal to 1.

Arrangement Problems

Arrangement problems, like the one involving our nine books, focus on counting the number of ways objects can be ordered or arranged. The key aspect of arrangement problems is that the order in which the objects appear matters. This is why permutation, not combination, is relevant here. Such problems are commonly found in everyday situations like arranging books on a shelf or ordering players in a lineup. To solve these, one must identify whether the order of selection impacts the final outcome. If order matters, we must use permutation formulas to determine the correct number of possible arrangements. In the given exercise, we calculated how many different ways we can arrange nine different books, signifying an arrangement problem.

• Arrangement problems consider the specific order of arranging objects.
• They are solved using permutation, as the sequence of objects is vital.
• Examples include scheduling, seating arrangements, and lineups.

Permutation Formula

Permutations deal with the arrangement of objects where the order is of utmost importance. The permutation formula is expressed as \( P(n,r) = \frac{n!}{(n-r)!} \), where \( n \) is the total number of items to choose from, and \( r \) is the number of items to arrange. This formula calculates the number of possible arrangements of \( r \) objects selected from \( n \) total objects.In the exercise at hand, it involves ensuring that the order in which we arrange the nine books is specific and counts. Therefore, we use the permutation formula with \( n = 9 \) and \( r = 9 \), simplifying to \( P(9,9) = \frac{9!}{(9-9)!} = \frac{9!}{0!} \), which equals 362,880.

• Permutations are used when order matters in the arrangement.
• Use the formula \( P(n,r) = \frac{n!}{(n-r)!} \) for calculations.
• The formula calculates possible sequences by considering the arrangement of objects.