Problem 51 - A Graphical Approach to Precalculus with Limits

Question

Solve each problem involving combinations. \(A\) banker's association has 30 members. If 4 members are selected at random to present a seminar, how many different groups of 4 are possible?

Step-by-Step Solution

Answer

There are 27,375 different groups possible.

Step 1: Understanding the Problem

The problem asks for the number of different ways to select 4 members from a group of 30. This is a classic combination problem where order does not matter.

Step 2: Define the Combination Formula

The formula to calculate combinations is given by \( C(n, k) = \frac{n!}{k!(n-k)!} \), where \( n \) is the total number of items to choose from, and \( k \) is the number of items to choose.

Step 3: Substitute Values into the Formula

Here, \( n = 30 \) and \( k = 4 \). Substitute these values into the combination formula: \[C(30, 4) = \frac{30!}{4!(30-4)!}\]

Step 4: Simplify Factorials

Calculate \( 30! \) only up until the 27th term since it will cancel with \( 26! \) in the denominator. Thus: \[\frac{30 \times 29 \times 28 \times 27}{4 \times 3 \times 2 \times 1}\]

Step 5: Perform the Calculations

First, calculate the numerator: \( 30 \times 29 \times 28 \times 27 = 657720 \).Next, calculate the denominator: \( 4 \times 3 \times 2 \times 1 = 24 \). Finally, divide the results: \( \frac{657720}{24} = 27375 \).

Step 6: Conclusion

The number of different groups of 4 members that can be selected from 30 members is 27,375.

Key Concepts

Factorials

The concept of factorials is central to understanding combinations and permutations. In mathematics, a factorial is denoted by an exclamation mark (!). It represents the product of all positive integers less than or equal to a given positive integer.

For example, the factorial of 3 is written as 3! and is calculated as:

• 3! = 3 × 2 × 1 = 6

Factorials are frequently used in combinatorial problems, as they help in counting the possible arrangements or selections of objects.
Remember that 0! is defined to be 1, which might seem a bit odd, but it fits with the rules of permutations and combinations by ensuring that calculations work out smoothly.

In our example, when calculating the combination of selecting 4 members from a group of 30, the factorials express the number of ways to arrange or select these members. The expression for this calculation, \( \frac{30!}{4!(26!)} \), uses factorial to reduce the complexity of these large numbers.

Combination Formula

The combination formula is a key concept in counting combinations, where order does not matter. For a set of items, the formula helps to find out how many different selections can be made. It's expressed as:

• \( C(n, k) = \frac{n!}{k!(n-k)!} \)

Where:

• \(n\) is the total number of items.
• \(k\) is the number of items to be chosen.

The importance of this formula lies in its ability to eliminate the need to manually count combinations, especially when dealing with large numbers of items. This comes handy in problems where the order of selection is irrelevant.

In the provided exercise of selecting 4 members from 30, the combination formula simplifies the counting process to yield:

• \( C(30, 4) = \frac{30!}{4!(26!)} = 27375 \)

This means there are 27,375 unique ways to choose 4 members from a group of 30, considering that the order of selection does not affect the outcome.

Permutation vs Combination

Understanding the distinction between permutations and combinations is crucial in solving problems related to arrangements and selections. Both concepts involve choosing from a set of items, but the importance of order differentiates them.

**Permutations** are used when the order of selection matters. If you are arranging books on a shelf, for instance, the order in which the books are placed affects the arrangement. The number of permutations is calculated using the formula:

• \( P(n, k) = \frac{n!}{(n-k)!} \)

**Combinations**, on the other hand, are used when the order does not matter. For example, selecting committee members from a larger group doesn't depend on the order in which they are picked. This is where the combination formula, discussed earlier, comes into play.

In our exercise, selecting 4 members out of 30 is a combination problem because the sequence in which the members are selected doesn’t change the group composition. Hence, the focus is solely on the number of ways to choose, not on how the chosen individuals are ordered within the group.