Problem 39 - Calculus for Biology and Medicine

Question

A collection contains seeds for four different annual and three different perennial plants. You plan a garden bed with three different plants, and you want to include at least one perennial. How many different selections can you make?

Step-by-Step Solution

Answer

31 different selections.

Step 1: Identify Total Combinations without Constraint

First, calculate the total number of ways to choose any 3 plants from the 7 available plants: 4 annuals and 3 perennials. Use the combination formula \(\binom{n}{k}\), where \(n\) is the total number of plants and \(k\) is the number of plants to choose: \(\binom{7}{3}\). Calculate this as follows: \[\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35\]. So, there are 35 ways to select any 3 plants from the 7.

Step 2: Calculate Combinations with No Perennials

Next, calculate the number of ways to select 3 plants that are all annuals. Since there are only 4 annual plants, use the combination formula \(\binom{4}{3}\). Calculate this as follows: \[\binom{4}{3} = \frac{4 \times 3 \times 2}{3 \times 2 \times 1} = 4\]. There are 4 ways to select 3 annual plants from the 4 available.

Step 3: Subtract Combinations without Perennials

Subtract the number of all-annual plant combinations from the total combinations to find the number of combinations that include at least one perennial. This is done by subtracting the all-annual combinations from the total combinations. Thus: \(35 - 4 = 31\). Therefore, there are 31 different ways to choose a combination of 3 plants that includes at least one perennial.

Key Concepts

Permutation and Combination

When planning how to select different plants for your garden, it's important to understand the difference between permutations and combinations.

- **Permutations** are all about arrangement. If you care about the order in which you choose your plants, you’d use permutations. But since order doesn't matter in this scenario, permutations aren't applicable here.- **Combinations**, on the other hand, are about choosing items without regard to order. In our garden planning, we use combinations because we just care about which plants are selected, not the sequence.

To calculate combinations, we use the formula \(inom{n}{k}\), which represents the number of ways to select **k** items from **n** items without regard to order. For our exercise, we calculated the different ways of choosing any 3 plants out of 7 total plants. This is computed as \(inom{7}{3} = 35\). This is the total selection of plants without considering any specific constraints.

Binomial Coefficient

The binomial coefficient is a fundamental concept when calculating combinations. It is often represented as \(inom{n}{k}\), meaning "n choose k." This shows how many ways you can choose **k** items from **n** total items. It is a key part of combinatorics.- The **binomial coefficient** formula is given by: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]- In our garden problem, the binomial coefficient \(inom{7}{3}\) was used to determine that there are 35 different ways to select any 3 plants.While the formula \[\binom{4}{3} = 4 \]helped us calculate combinations without perennials, allowing us to then focus on ensuring at least one perennial was included. Understanding how to manipulate these coefficients is crucial in solving probability and combinatorial problems effectively.

Constraint Handling

Constraint handling in combinatorics often requires you to adjust calculations based on specific limitations or requirements. In the garden problem, the key constraint was that the selection must include at least one perennial plant.

To manage this, we first looked at the total combinations without constraints and then subtracted those combinations that didn't meet the perennial requirement.

• First, all potential combinations with any plants were calculated as \(inom{7}{3} = 35\).
• Next, we found there were \(inom{4}{3} = 4\) combinations using only annuals.
• To address the constraint of including at least one perennial, we subtracted these all-annuals combinations from the total.

This subtraction was \(35 - 4 = 31\), ensuring we only counted combinations that include at least one perennial. This strategy can be applied to various scenarios where specific conditions must be met, making constraint handling an essential skill in combinatorial calculations.