Problem 32 - Calculus for Biology and Medicine

Question

A bag contains 45 beans of three different varieties. Each variety is represented 15 times in the bag. You grab 9 beans out of the bag. (a) Count the number of ways that each variety can be represented exactly three times in your sample. (b) Count the number of ways that only one variety appears in your sample.

Step-by-Step Solution

Answer

(a) 455^3 ways for each variety; (b) 15,015 ways for one variety.

Step 1: Understand the Problem

We are given a bag containing 45 beans, with three varieties of beans (A, B, and C), each having 15 beans. We need to draw 9 beans and determine the number of ways each variety can be represented exactly three times or if only one variety appears in the sample.

Step 2: Representing Each Variety Three Times

We want to choose 3 beans from each of the 3 varieties. Since there are 15 beans of each variety, the number of ways to choose 3 beans from a variety is given by the combination formula: \( \binom{n}{k} \). Apply this to each variety: \( \binom{15}{3} \).

Step 3: Calculate Combinations for Each Variety

For each variety, calculate \( \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \). Since there are three varieties, we'll have \( 455^3 \) ways to choose 3 beans from each variety.

Step 4: Calculate Total Ways for Equal Representation

Multiply the ways to choose from each variety together: \( 455 \times 455 \times 455 \). So, \( 455^3 \) ways. This gives the total ways of choosing 9 beans with each type represented three times.

Step 5: One Variety in the Sample

To have only one variety in the sample, all 9 beans must be of the same type. Since we only have 15 beans of each type, choose all 9 from one type: \( \binom{15}{9} \). This is calculated as: \( \binom{15}{9} = \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7}{9!} \).

Step 6: Calculate Combinations for One Variety

By calculating the above, \( \binom{15}{9} = 5005 \) for one variety. Since there are three varieties that could be chosen, multiply by 3: \( 3 \times 5005 = 15015 \).

Step 7: Conclusions

For part (a), the number of ways each variety can be represented exactly three times is \( 455^3 \). For part (b), the number of ways only one variety appears is 15,015.

Key Concepts

Binomial Coefficient

The binomial coefficient is a fundamental concept in combinatorics, and it represents the number of ways to choose a subset of elements from a larger set. In mathematical terms, it's expressed as \( \binom{n}{k} \), which means "n choose k." This formula is crucial for counting combinations, where order doesn't matter.

• The formula is \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \(!\) denotes factorial, meaning the product of all positive integers up to that number.
• For example, in our exercise, we used \( \binom{15}{3} \) to calculate how to choose 3 beans from 15, resulting in 455 ways.
• This calculation assumes each variety of beans is treated independently.

Understanding the binomial coefficient can simplify many problems related to grouping and selections. It's useful not only in combinatorics but also in probability and various fields of discrete mathematics.

Probability

Probability is the measure of the likelihood that an event will occur. In this exercise, knowing how to calculate combinations is part of assessing probabilities.

• When calculating how each variety can be represented three times or just one variety entirely, we're considering different probable outcomes based on the method of selecting the beans.
• Using probabilities, you can determine the chance of drawing any specific configuration (like all 9 beans being of one variety).
• Probability helps in understanding chances in various discrete mathematical problems and can guide decision-making processes under uncertainty.

By honing our understanding of probability in this context, we can better appreciate the role of randomness and choice within combinatorial spaces.

Discrete Mathematics

Discrete mathematics includes the study of distinct and separate objects. It's the mathematics of countable substances and has connections to combinatorics.

• This field includes topics like logic, set theory, graph theory, and of course, combinatorics.
• Problems like our bean selection exercise are inherently discrete because you deal with finite, countable items (beans).
• Discrete mathematics provides the tools to tackle various problems involving integers, finite structures, and can be used in algorithms and computer science applications.

Mastering discrete mathematics concepts will provide a foundation not just for academics but for real-world applications in technology and engineering.