Problem 29 - A Graphical Approach to College Algebra

Question

Use a calculator to evaluate each expression. $$_{20} C_{5}$$

Step-by-Step Solution

Answer

There are 15,504 combinations.

Step 1: Understand the Notation

The notation \( _{20}C_5 \) represents a combination, which is the number of ways to choose 5 objects from a set of 20, where the order does not matter. This is calculated using the combination formula: \[ _nC_r = \frac{n!}{r!(n-r)!} \] where \( n = 20 \) and \( r = 5 \).

Step 2: Calculate Factorials

Calculate the factorials of the relevant numbers. - \( 20! \) is the product of all integers from 20 to 1. - \( 5! \) is the product of all integers from 5 to 1. - \( (20-5)! = 15! \) is the product of all integers from 15 to 1. Using a calculator, find these values:- \( 20! = 2,432,902,008,176,640,000 \)- \( 5! = 120 \)- \( 15! = 1,307,674,368,000 \).

Step 3: Apply the Combination Formula

Now, apply the combination formula: \[ _{20}C_5 = \frac{20!}{5!(20-5)!} = \frac{20!}{5! \times 15!} \] Substitute the factorial values: \[ _{20}C_5 = \frac{2,432,902,008,176,640,000}{120 \times 1,307,674,368,000} \]

Step 4: Solve the Expression

Compute the division to find the result: \[ _{20}C_5 = \frac{2,432,902,008,176,640,000}{157,216,544,160,000} = 15,504 \] Use it shows there are 15,504 ways to choose 5 objects from a set of 20 without considering the order.

Key Concepts

Factorials

Factorials are an essential building block in the world of combinatorics and various mathematical calculations. They are denoted with an exclamation mark, such as \( n! \). This notation signifies the product of all positive integers up to \( n \).
Let’s break down how factorials work:

• \( 0! \) is defined to be 1 by convention.
• \( 1! \) is simply 1, since there’s only one number to multiply.
• \( 2! \) is calculated as \( 2 \times 1 = 2 \).
• Extending this pattern, \( 5! \) would be \( 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

Factorials grow rapidly with increasing \( n \). For instance, \( 20! \) is a substantially large number: 2,432,902,008,176,640,000. Calculators are often needed to handle these computations, especially when dealing with such large values.
Furthermore, factorials are ubiquitous in various mathematical formulas, particularly in permutations and combinations.

Combination Formula

The combination formula provides a method to calculate the number of ways to select \( r \) objects from a pool of \( n \) objects, where the order of selection does not matter. This formula is vital in scenarios where arrangement or sequence isn't considered important.
The formula is expressed as:\[ _nC_r = \frac{n!}{r!(n-r)!} \]

• \( n \) is the total number of objects.
• \( r \) is the number of objects to choose.
• The denominator \( r!(n-r)! \) accounts for the different orders that can occur.

Understanding this formula is crucial, as it enables you to tackle problems that involve the selection of items in fields such as probability and statistics. By substituting the values of \( n = 20 \) and \( r = 5 \) into the formula, we can calculate \( _{20}C_5 \) by determining the relevant factorials, as shown in the exercise.

Binomial Coefficient

The binomial coefficient is a central concept in combinatorics. It is often represented as \( _nC_r \) or \( \binom{n}{r} \), and equates to the number of combinations or ways to choose \( r \) elements from a set of \( n \) elements, disregarding the order of those elements.
This is what we solve when using the combination formula. The binomial coefficient appears prominently not just in combinatorics, but also in the binomial theorem, which relates to expanding expressions that are raised to a power.

• The binomial theorem uses these coefficients to display each expanded term.
• It's applicable in polynomial expansions and understanding pascals triangle.

In practice, once we have calculated the factorial values, solving for a binomial coefficient like \( _{20}C_5 \) allows us to understand in how many different ways a subset can be formed from the larger set, which in this case results in 15,504 different ways.