Problem 45 - Algebra and Trigonometry

Question

These problems involve combinations. Choosing Books In how many ways can three books be chosen from a group of six different books?

Step-by-Step Solution

Answer

There are 20 ways to choose 3 books from 6 different books.

Step 1: Understand the problem

We need to find out how many ways we can choose 3 books from 6 different books. This is a combinations problem where order does not matter.

Step 2: Identify the formula

For combinations, we use the formula \( _nC_r = \frac{n!}{r!(n-r)!} \), where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose.

Step 3: Plug in the numbers

Here, \( n = 6 \) and \( r = 3 \). So, substitute these values into the formula: \( _6C_3 = \frac{6!}{3!(6-3)!} \).

Step 4: Simplify the factorials

Calculate the factorials: \( 6! = 720 \), \( 3! = 6 \), and \( (6-3)! = 3! = 6 \).

Step 5: Perform the calculation

Substitute the factorials into the formula: \[ _6C_3 = \frac{720}{6 \times 6} = \frac{720}{36} = 20 \]. Thus, there are 20 ways to choose the books.

Key Concepts

Factorials

Factorials are a fundamental concept in mathematics, especially when dealing with probability and combinatorics. The factorial of a non-negative integer, denoted by an exclamation point (!), is the product of all positive integers less than or equal to that number. For example, the factorial of 4, written as 4!, is calculated as follows:

• 4! = 4 × 3 × 2 × 1 = 24

Factorials grow very quickly. Even small numbers can lead to large results. For instance, 5! is 120, and 6! is a substantial 720. Understanding factorials is crucial when solving combination or permutation problems because these operations fundamentally rely on their usage.


The formula for combinations, which is often used in these types of math problems, incorporates factorials to help determine different groupings and arrangements. This makes it easier to calculate how many ways you can organize or choose items, like choosing 3 books out of 6 in our example problem.

Combinatorial Mathematics

Combinatorial mathematics is a field that focuses on counting, arranging, and analyzing combinations and permutations of sets. Combinatorial methods are widely applied in computer science, statistics, and other disciplines where decision structures are considered. A basic concept is understanding combinations, which are selections of items from a larger set where order does not matter.

When discussing combinations, the formula \[_nC_r = \frac{n!}{r!(n-r)!} \]comes into play. Here, \(n\) represents the total number of items to choose from, and \(r\) indicates how many items we are selecting. This formula tells us the number of possible combinations and exemplifies the beauty and efficiency of combinatorial mathematics.

In our original exercise of choosing 3 books from 6, we use the concept of combinations. By plugging the numbers into our formula, we can systematically determine the number of different ways to select the books without regard to order, which in this case, results in 20 combinations.

Permutations

Permutations are similar to combinations, but with a key difference: order matters. A permutation refers to an arrangement of all or part of a set of objects, with regard to the order of the arrangement. The formula for permutations generally involves factorials and is more complex than combinations.

For a permutation problem, where order is important, the number of permutations of \( n \) items taken \( r \) at a time is given by: \[_nP_r = \frac{n!}{(n-r)!}\]This highlights the permutations as being more numerous than combinations because each distinct order counts as a separate arrangement.

Understanding when to use permutations or combinations is critical based on whether order influences the arrangement. In our exercise, using combinations was appropriate because the order of the books does not matter when selecting which ones to choose. However, in a scenario where the sequence of choices is important, permutations would be the required approach.