Problem 24 - Calculus for Biology and Medicine

Question

Twelve people wait in front of an elevator that has room for only \(5 .\) Count the number of ways that the first group of people to take the elevator can be chosen.

Step-by-Step Solution

Answer

There are 792 ways to choose 5 people from a group of 12 to take the elevator.

Step 1: Understanding Combinations

When the order of selection does not matter, such as choosing people to get into an elevator, we use combinatorics to find the number of ways to select a group. Specifically, we use combinations, denoted as \( \binom{n}{k} \), which represents the number of ways to choose \( k \) people from a group of \( n \).

Step 2: Apply Combination Formula

We need to select 5 people from a total of 12. The formula for combinations is \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \). Here, \( n = 12 \) and \( k = 5 \). Plug these into the formula: \[ \binom{12}{5} = \frac{12!}{5!(12-5)!} \]

Step 3: Calculate the Factorials

Calculate the factorials as follows: - \( 12! = 12 \times 11 \times 10 \times 9 \times 8 \times 7! \) - \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \) - \( 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \)

Step 4: Simplify the Combination Expression

Now, substitute the factorial values into our combination formula:\[\binom{12}{5} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7!}{5! \times 7!}\]The \( 7! \) cancels out, leaving:\[\binom{12}{5} = \frac{12 \times 11 \times 10 \times 9 \times 8}{120}\]

Step 5: Compute the Result

Calculate the multiplication in the numerator:\[ 12 \times 11 \times 10 \times 9 \times 8 = 95040 \]Now divide by 120:\[ \frac{95040}{120} = 792 \]Thus, there are 792 different ways to choose 5 people from 12 to go into the elevator.

Key Concepts

Combinations

Combinations are a fundamental concept in combinatorics where the order of selection does not matter. Imagine you have a group of people, and you need to select a smaller group to perform a task, like getting into an elevator. The crucial aspect of combinations is that each selected group is treated the same, regardless of how its members were chosen. For example, selecting Alice, Bob, and Carol is the same as selecting Carol, Alice, and Bob.

To calculate combinations, we use the formula:

• \( \binom{n}{k} \), where \( n \) is the total number of items to choose from, and \( k \) is the number of items to choose.
• The formula for \( \binom{n}{k} \) is \( \frac{n!}{k!(n-k)!} \).

In our example with the elevator, we had 12 people (\( n = 12 \)) and needed to select 5 (\( k = 5 \)). By applying the combination formula, we simplify the selection process among groups.

Factorial Calculation

The factorial calculation is a key part of understanding permutations and combinations. A factorial, denoted by the symbol \( ! \), is the product of all positive integers up to a specified number. For instance, \( n! \) means multiplying all whole numbers from 1 to \( n \).

Here are some examples to illustrate the calculation:

• \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
• \( 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \)

In the elevator problem, factorials help in simplifying the combination formula. We focus on dividing \( n! \) by the products of \( k! \) and \((n-k)! \).

This removal of common terms helps reduce complexity and simplify calculations.

Permutations

Permutations differ from combinations in that they consider the order of selection. If the sequence of selected items matters, we are dealing with permutations rather than combinations. For example, if picking a specific order for people to speak matters, then we're interested in permutations.

The formula for permutations is written as:

• \( P(n, k) = \frac{n!}{(n-k)!} \)

This formula calculates how many ways you can arrange \( k \) items out of a total \( n \) where the order is important.

While permutations are not directly used in the original elevator problem, understanding the distinction is helpful for broader combinatorial problems. Keeping track of when order matters can help decide if you're calculating a permutation or a combination.