Problem 35 - College Algebra

Question

Use the formula for \(_{n} P_{r}\) to solve Exercises \(33-40\) For a segment of a radio show, a disc jockey can play 7 records. If there are 13 records to select from, in how many ways can the program for this segment be arranged?

Step-by-Step Solution

Answer

The program for this segment can be arranged in 19,958,400 ways.

Step 1: Identify \(n\) and \(r\)

We know from the problem that the number of total records (n) is 13, and our disc jockey can play 7 records (r). So, \(n=13\) and \(r=7\).

Step 2: Substitute \(n\) and \(r\) Into the Formula

Plug the values of \(n\) and \(r\) into the formula \(_{n} P_{r} = \frac{n!}{(n-r)!}\). So, we have \(_{13} P_{7} = \frac{13!}{(13-7)!}\).

Step 3: Simplify the formula

Subtract in the denominator \(13 - 7 = 6\). The new formula becomes \(_{13} P_{7} = \frac{13!}{6!}\).

Step 4: Calculate the factorial and evaluate the fraction

13 factorial (13!) is \(13*12*11*10*9*8*7*6*5*4*3*2*1\) and 6 factorial (6!) is \(6*5*4*3*2*1\). Divide \(13!\) by \(6!\). Because of the common terms in the numerator and denominator, we can cancel the terms from 6 to 1 in both numerator and denominator which simplifies to \(13*12*11*10*9*8*7\). Calculate this product to give a final answer.

Key Concepts

Understanding Factorials

A factorial, denoted by an exclamation point (!), is a fundamental concept in permutations, where it represents the product of all positive integers up to a certain number. This product helps calculate the total number of arrangements or orderings for a set number of items.
For instance, the factorial of 5, written as 5!, is the product of all positive integers less than or equal to 5. In mathematical terms, it's:

• 5! = 5 × 4 × 3 × 2 × 1 = 120

The concept of factorial is significant because, in permutation problems, it allows us to account for all the possible reordering of objects. When you see a factorial in a problem, remember it's just a way of organizing or arranging items in multiple ways.
Factorials grow rapidly; that is, as the number increases, the factorial becomes substantially larger. Being able to simplify and calculate these efficiently is important when dealing with permutations.

Combinations vs. Permutations

Both combinations and permutations are about selecting items from a larger set. However, the fundamental difference between the two lies in the importance of order.
In permutations, the order matters. For example, selecting the first-place, second-place, and third-place winners in a race means that the order in which they are arranged is significant. Combinations, on the other hand, do not consider order; they merely care about the selection of items.
Here's a simplified way to remember:

• Permutation: Order matters. Arranging books on a shelf, creating passwords, etc.
• Combination: Order doesn’t matter. Selecting a committee from a group, picking lottery numbers, etc.

Knowing when to use permutations versus combinations is crucial in solving mathematical problems correctly. The permutation formula and the combination formula will vary slightly to reflect whether or not order is a factor. For permutations, the formula is:

• \[ _{n} P_{r} = \frac{n!}{(n-r)!} \]

Where \(n\) is the total number of items, and \(r\) is the number of selections.

The Permutation Formula Explained

The permutation formula is used to find the number of different ways you can arrange a specific set of items. The formula is relatively straightforward but immensely powerful, especially in scenarios like scheduling, organizing events, or even arranging books.
To apply the permutation formula, use:

• \[ _{n} P_{r} = \frac{n!}{(n-r)!} \]

Where:

• \(n\) = total number of items to choose from.
• \(r\) = number of items being arranged or selected.

In practice, take for example, a scenario with 13 records, where only 7 need to be arranged for a radio show. Here it is crucial to consider that even with the same records, swapping two would change the arrangement altogether, highlighting the need for permutations.
The main steps involve calculating \(n!\) or the total arrangements for all items, and then dividing it by the arrangements of the items we're not using \((n-r)!\). This way, you're left with just the arrangements of the chosen items.
This formula highlights the power of order. Without considering different sequences, you can't fully gauge the possibilities. Understanding this formula can open up new perspectives on issues involving arrangement and order.