Finding the Number of Subsets with Exactly Two Elements

The question asks us to find the number of subsets of a given set A that contain exactly two elements. The set A is defined as A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

The set A has a total of 10 elements. We are interested in forming subsets of A that have a specific size, which is exactly 2 elements.

When we form a subset, the order of the elements does not matter. For example, the subset {1, 2} is the same as the subset {2, 1}. This means we are choosing a group of 2 elements from the 10 available elements without considering the order. This type of problem involves combinations.

The number of ways to choose $k$ elements from a set of $n$ distinct elements, where the order does not matter, is given by the combination formula:

\( C(n, k) = \binom{n}{k} = \frac{n!}{k!(n-k)!} \)

In this problem:

• \( n \) is the total number of elements in the set A, which is 10.
• \( k \) is the number of elements we want in each subset, which is 2.

So, we need to calculate the number of combinations of choosing 2 elements from 10 elements, which is \( C(10, 2) \).

Let's calculate \( C(10, 2) \) using the formula:

\( C(10, 2) = \frac{10!}{2!(10-2)!} \)

\( C(10, 2) = \frac{10!}{2!8!} \)

Now, let's expand the factorials:

\( 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \)

\( 2! = 2 \times 1 = 2 \)

\( 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \)

Substitute these into the formula:

\( C(10, 2) = \frac{10 \times 9 \times 8!}{2 \times 1 \times 8!} \)

We can cancel out \( 8! \) from the numerator and the denominator:

\( C(10, 2) = \frac{10 \times 9}{2 \times 1} \)

\( C(10, 2) = \frac{90}{2} \)

\( C(10, 2) = 45 \)

Thus, there are 45 different subsets of A that contain exactly two elements.

Revision Table: Key Concepts

Additional Information: Set Theory and Counting

Understanding sets and counting principles like combinations and permutations is fundamental in mathematics, especially in probability and statistics. Here's a bit more information:

• Permutations vs. Combinations: Permutations are used when the order of selection matters (e.g., arranging books on a shelf), while combinations are used when order doesn't matter (e.g., selecting a committee). The formula for permutations is \( P(n, k) = \frac{n!}{(n-k)!} \).
• Total Number of Subsets: A set with \( n \) elements has \( 2^n \) total subsets, including the empty set and the set itself. For set A with 10 elements, the total number of subsets is \( 2^{10} = 1024 \).
• Subsets of Specific Size: The number of subsets of size \( k \) from a set of size \( n \) is given exactly by the combination formula \( \binom{n}{k} \), as used in this problem. The sum of the number of subsets of each possible size equals the total number of subsets: \( \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + ... + \binom{n}{n} = 2^n \).
• Empty Set and Set A: The number of subsets with 0 elements is \( \binom{10}{0} = 1 \). This is the empty set \( \{\} \). The number of subsets with 10 elements is \( \binom{10}{10} = 1 \). This is the set A itself.

This problem specifically focused on finding subsets of a particular size using combinations, which is a common application of combinatorics in set theory.