Finding Subsets of a Set with Specific Sizes

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

The total number of elements in set A is 10. We need to find the number of subsets of A with a specific number of elements. This is a problem related to combinations in combinatorics.

The number of ways to choose \(k\) elements from a set of \(n\) distinct elements, without regard to the order of selection, is given by the combination formula:

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

Here, \(n\) is the total number of elements in the set A, which is 10. We are interested in two cases:

1. Subsets containing two elements (\(k=2\)).
2. Subsets containing three elements (\(k=3\)).

Number of Subsets with Two Elements

To find the number of subsets of A with two elements, we need to calculate \(\binom{10}{2}\). Using the combination formula with \(n=10\) and \(k=2\):

$$ \binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10!}{2!8!} $$

Expanding the factorials:

$$ \binom{10}{2} = \frac{10 \times 9 \times 8!}{ (2 \times 1) \times 8! } $$

Cancel out \(8!\) from the numerator and denominator:

$$ \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = \frac{90}{2} = 45 $$

So, there are 45 subsets of A containing exactly two elements.

Number of Subsets with Three Elements

To find the number of subsets of A with three elements, we need to calculate \(\binom{10}{3}\). Using the combination formula with \(n=10\) and \(k=3\):

$$ \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!} $$

Expanding the factorials:

$$ \binom{10}{3} = \frac{10 \times 9 \times 8 \times 7!}{ (3 \times 2 \times 1) \times 7! } $$

Cancel out \(7!\) from the numerator and denominator:

$$ \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120 $$

So, there are 120 subsets of A containing exactly three elements.

Total Number of Subsets with Two or Three Elements

The question asks for the number of subsets containing two or three elements. Since a subset cannot simultaneously contain exactly two elements and exactly three elements, these two cases are mutually exclusive. Therefore, we can find the total number by adding the number of subsets with two elements and the number of subsets with three elements.

Total number of subsets = (Number of subsets with two elements) + (Number of subsets with three elements)

Total number of subsets = \(\binom{10}{2} + \binom{10}{3}\)

Total number of subsets = \(45 + 120 = 165\)

Summary of Calculations

The total number of subsets of A containing two or three elements is 165.

Revision Table: Combinations and Subsets

Additional Information: Combinatorics Basics

Combinatorics is a branch of mathematics dealing with combinations and permutations. It is used for counting arrangements and selections of objects.

• Permutation: An arrangement of objects in a specific order. The number of permutations of choosing \(k\) objects from \(n\) is \(P(n, k) = \frac{n!}{(n-k)!}\). Order matters in permutations.
• Combination: A selection of objects where the order does not matter. Subsets are examples of combinations because the order of elements within a set does not change the set itself (e.g., {1, 2} is the same set as {2, 1}).
• In this problem, we are counting subsets of a specific size. Since the order of elements within a subset doesn't matter, we use combinations (\(\binom{n}{k}\)).
• The total number of subsets of a set with \(n\) elements is \(2^n\). This is because for each element, there are two choices: either it is in the subset or it is not.
• The sum of the number of subsets of all possible sizes (from 0 to n) equals the total number of subsets: \(\sum_{k=0}^{n} \binom{n}{k} = 2^n\).
• For set A with 10 elements, the total number of subsets is \(2^{10} = 1024\). This total includes subsets of size 0, 1, 2, ..., 10. Our problem only considers subsets of size 2 and 3.