Problem 4 - Algebra and Trigonometry

Question

True or false? If we have a set with five distinct objects, then the number of different ways of choosing two members of this set is the same as the number of ways of choosing three members.

Step-by-Step Solution

Answer

True, both are equal to 10.

Step 1: Define the Problem

We need to determine if the number of ways to choose two members from a set of five is equal to the number of ways to choose three members from the same set.

Step 2: Use Combinatorial Formula for Selections

The number of ways to choose \( k \) members from a set of \( n \) members is given by the combination formula: \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \).

Step 3: Calculate Combinations for Two Members

We use the formula \( \binom{n}{k} = \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \) ways.

Step 4: Calculate Combinations for Three Members

Similarly, calculate the combinations for three members: \( \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 \) ways.

Step 5: Compare the Results

We found that \( \binom{5}{2} = 10 \) and \( \binom{5}{3} = 10 \). Since both results are equal, the statement is true.

Key Concepts

Combination Formula

The combination formula is an essential tool in combinatorics. It allows us to find the number of ways to choose a subset of items from a larger set without considering the order of selection.
In mathematical terms, if we have a set of size \( n \) and want to choose \( k \) items, the formula is: \[\binom{n}{k} = \frac{n!}{k!(n-k)!} \]Here, \( n! \) (read as 'n factorial') signifies the product of all positive integers up to \( n \), while \( k! \) and \((n-k)!\) are the factorials of \( k \) and \( n-k \) respectively.
This formula simplifies the process of calculating combinations by accounting for all possible arrangements and then adjusting for repeated sequences (since order doesn't matter in combinations).

• The order of selection is not important.
• Factorials play an integral role in calculating combinations.

Notice how the formula is symmetric: choosing \( k \) items from \( n \) items leaves \( n - k \) items unchosen, which means that \( \binom{n}{k} = \binom{n}{n-k} \). This symmetry is what allows us to say in our example that choosing 2 members is equivalent to choosing 3 from 5 members because \( \binom{5}{2} = \binom{5}{3} \).

Set Theory

Set theory is a fundamental part of modern mathematics, dealing with the collection of objects, known as sets. It provides a solid framework for defining and understanding various mathematical concepts, including combinations.
A set is essentially a collection of distinct objects. The objects in the set are called members or elements.

• Order of elements doesn't matter in a set.
• Each element is unique within the set.

For the exercise mentioned, we had a set of five distinct objects. When determining the number of ways to choose members from a set, the concept of distinctness in set theory is crucial. This means that no two elements in our set are the same, and every choice consists solely of these unique elements.
In our example, the set consisted of 5 elements, allowing us to explore the number of potential groups formed, considering the unique nature of each element. Set theory simplifies our understanding by focusing on the nature of element choice without the complications of ordering, which is preserved in our combination calculations.

Factorials

Factorials are a key mathematical concept often used to calculate permutations and combinations. The factorial of a non-negative integer \( n \), denoted as \( n! \), is the product of all positive integers less than or equal to \( n \).

• \( n! = n \times (n-1) \times (n-2) \times \ldots \times 1 \).
• The factorial of 0 is defined as 1, that is, \( 0! = 1 \).

Factorials are particularly useful in permutations where order matters, but they become equally important in combinations when accounting for repeated patterns. They help to calculate the total number of potential orderings and adjust for those we aren't considering (as in combinations).
In our exercise, this was evident when we calculated the number of combinations for choosing 2 and 3 members from 5 elements. Factorials provided the groundwork for the combination formula, simplifying the computation and making it manageable:
\( \binom{5}{2} = \frac{5!}{2!\times3!} = \frac{120}{12} = 10 \).
Thus, understanding factorials is fundamental in mastering the use of the combination formula as well as other combinatorial problems.