Problem 62 - A Graphical Approach to College Algebra

Question

Use any or all of the methods described in this section to solve each problem. Pineapple Samples How many samples of 3 pineapples can be drawn from a crate of \(12 ?\)

Step-by-Step Solution

Answer

There are 220 ways to draw 3 pineapples from a crate of 12.

Step 1: Identify the problem as a combination

We need to determine the number of ways to choose 3 pineapples from a total of 12 pineapples. This is a problem of combinations, where the order of choosing does not matter, and we are looking to find \( \binom{12}{3} \).

Step 2: Use the combination formula

The formula for combinations is given by \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \). Here, \( n = 12 \) and \( r = 3 \). We will plug these values into the formula.

Step 3: Calculate factorial values

First, calculate the factorial values needed: \( 12! \), \( 3! \), and \( (12-3)! = 9! \). We need:- \( 3! = 3 \times 2 \times 1 = 6 \)- \( 9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362880 \)

Step 4: Simplify the combination formula

Substitute the factorial values into the combination formula:\[ \binom{12}{3} = \frac{12!}{3! \times 9!} = \frac{12 \times 11 \times 10 \times 9!}{3! \times 9!} \] We can cancel \( 9! \): \[ \frac{12 \times 11 \times 10}{3!} = \frac{12 \times 11 \times 10}{6} \]

Step 5: Perform the division

Calculate \( \frac{12 \times 11 \times 10}{6} \):- \( 12 \times 11 = 132 \)- \( 132 \times 10 = 1320 \)- \( \frac{1320}{6} = 220 \) Thus, there are 220 different combinations of drawing 3 pineapples from 12.

Key Concepts

Factorial

Factorials are an essential concept in mathematics, especially in dealing with combinations and permutations. The factorial of a non-negative integer is the product of all positive integers less than or equal to that number. It's denoted by an exclamation point (!).
For instance, the factorial of 3 is written as \(3!\) and calculated as follows:

• \(3! = 3 \times 2 \times 1 = 6\)

Factorials grow very quickly with larger numbers, which is why they are often used in calculating combinations. For example,

• \(12!\) denotes the factorial of 12
• It is the product of integers from 1 to 12
• \(12! = 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 479001600\)

Factorials are a key part of the formulas used in combinations and permutations, helping us count the number of possible ways to arrange or select items.

Combination Formula

The combination formula is used when the order of selection does not matter. It helps to find how many ways you can choose \(r\) items from \(n\) items without regard for order. The formula itself is:

• \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\)

This formula calculates the number of combinations, where:

• \(n\) is the total number of items
• \(r\) is the number of items to choose

In the pineapple problem, we used this formula to calculate \(\binom{12}{3}\), meaning we selected 3 out of 12 pineapples:

• Plug in the values: \(\binom{12}{3} = \frac{12!}{3! \times (12-3)!} = \frac{12!}{3! \times 9!}\)
• Since \(9!\) is common in the numerator and denominator, it cancels out
• Simplifying further: \(\frac{12 \times 11 \times 10}{6}\)

Ultimately, the combination formula helps solve problems where order doesn't matter.

Permutations

Permutations are a mathematical concept where the order of items does indeed matter. Unlike combinations, permutations count the number of ways to arrange \(n\) items into a specific sequence. This makes permutations different from combinations where the specific order is not important.
The formula for permutations is given as:

• \(P(n, r) = \frac{n!}{(n-r)!}\)

Here,

• \(n\) is the total number of items
• \(r\) is the number of items being arranged

For instance, if we wanted to arrange 3 pineapples out of 12 in a specific order, we would find \(P(12, 3)\):

• Calculate using the permutation formula: \(P(12, 3) = \frac{12!}{(12-3)!}\)
• This is equivalent to \(\frac{12 \times 11 \times 10 \times 9!}{9!}\)
• After canceling \(9!\), we have \(12 \times 11 \times 10\)

Permutations result in a larger number than combinations when \(r\) is not equal to \(n\) because they account for different sequences of the same set of items.