Problem 23 - Calculus for Biology and Medicine

Question

An urn contains 15 different balls. In how many ways can you select 4 balls without replacement?

Step-by-Step Solution

Answer

There are 1365 ways to select 4 balls from 15 balls without replacement.

Step 1: Understanding the Problem

We need to determine the number of ways to select 4 balls from a total of 15 balls, where the order does not matter and each ball can only be selected once.

Step 2: Using the Combination Formula

Because the problem does not consider the order of selection and involves selecting without replacement, we use the combination formula. The formula for combinations is: \( C(n, r) = \frac{n!}{r!(n-r)!} \), where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose.

Step 3: Substituting Values into the Formula

Here, \( n = 15 \) and \( r = 4 \). Substitute these values into the combination formula: \[ C(15, 4) = \frac{15!}{4!(15-4)!} \].

Step 4: Simplifying the Factorials

Calculate the factorials in the formula: - \( 15! = 15 \times 14 \times 13 \times 12 \times 11! \)- \( 4! = 4 \times 3 \times 2 \times 1 = 24 \)- Thus, \( (15-4)! = 11! \). So, we simplify: \[ C(15, 4) = \frac{15 \times 14 \times 13 \times 12 \times 11!}{4! \times 11!} \].

Step 5: Canceling Out Factorials

Since \( 11! \) appears in both the numerator and the denominator, they cancel each other out, leaving: \[ C(15, 4) = \frac{15 \times 14 \times 13 \times 12}{24} \].

Step 6: Calculating the Final Answer

Now compute \( \frac{15 \times 14 \times 13 \times 12}{24} \).First calculate the numerator: - \( 15 \times 14 = 210 \)- \( 210 \times 13 = 2730 \)- \( 2730 \times 12 = 32760 \)Then divide by \( 24 \):- \( 32760 \div 24 = 1365 \).This gives us the final answer.

Key Concepts

Combination Formula

When selecting a subset from a larger set where the order does not matter, we often use the combination formula. This essential tool helps us in problems where we're picking items without caring about the sequence. In combinatorics, combinations are distinct from permutations, which do take order into account. The formula for determining combinations is:

• \[ C(n, r) = \frac{n!}{r!(n-r)!} \]

Here, \( n \) is the total number of items available, and \( r \) is the number of items we want to choose. The exclamation mark (!) represents a factorial, a concept we will explore next. By using this formula, we ensure accurate calculations when the selection order is irrelevant, making it a valuable asset in many statistical and mathematical applications.

Factorials

Factorials are a fundamental concept in combinatorics and beyond, frequently used in calculating combinations and permutations. A factorial of a number \( n \) (denoted as \( n! \)) is the product of all positive integers less than or equal to \( n \). For example:

• \( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
• \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)

Factorials represent the number of ways to arrange \( n \) objects in a sequence. In the context of the combination formula, using factorials helps find the total possible ways to choose a subset of items. They are crucial for simplifying expressions and canceling terms, like in our exercise where we canceled \( 11! \) from both the numerator and denominator. Understanding how factorials interact within the formula equips us to solve complex problems easily.

Counting Methods

In the vast realm of combinatorics, counting methods are tools that help us determine the number of ways to perform a selection or arrangement task. They are essential for calculating combinations and permutations. Two primary methods we often encounter are:

• **Combinations** - Applied when the order of selection does not matter. Used extensively in problems like selecting lottery numbers or forming a team from a group.
• **Permutations** - Used when the order of selection does matter. Common in tasks like determining seating arrangements or organizing a schedule.

The key difference lies in whether order influences the count. By categorizing your problem correctly, you can choose the right method and find accurate solutions. Counting methods enable you to tackle a diversity of mathematical challenges with confidence and precision.