How many subsets of cardinality 3 does A have if |A| = n?
Understand the Problem
The question is asking for the number of subsets of a set A that have exactly 3 elements, given that the cardinality (size) of set A is n. This typically involves combinatorial mathematics, specifically choosing 3 elements from a larger set of n elements.
Answer
The number of subsets of cardinality 3 is given by $$ \frac{n(n-1)(n-2)}{6} $$
Answer for screen readers
The number of subsets of cardinality 3 from set ( A ) is given by
$$ \binom{n}{3} = \frac{n(n-1)(n-2)}{6} $$
Steps to Solve
- Identify the problem's requirement
We need to find the number of subsets of a set ( A ) that contain exactly 3 elements. The size of set ( A ) is given as ( n ).
- Use the combination formula
To calculate how many ways we can choose 3 elements from ( n ) elements, we use the combination formula:
$$ \binom{n}{k} = \frac{n!}{k!(n-k)!} $$
In this scenario, ( k = 3 ).
- Set the values in the formula
Substituting ( k = 3 ) into the combination formula:
$$ \binom{n}{3} = \frac{n!}{3!(n-3)!} $$
- Simplify the expression
This can be simplified further by calculating ( 3! ) which equals 6:
$$ \binom{n}{3} = \frac{n(n-1)(n-2)}{3!} = \frac{n(n-1)(n-2)}{6} $$
The number of subsets of cardinality 3 from set ( A ) is given by
$$ \binom{n}{3} = \frac{n(n-1)(n-2)}{6} $$
More Information
This formula represents the number of ways to choose 3 items from ( n ) distinct items without regard to the order of selection. It is an essential concept in combinatorics and is commonly used in probability and statistical calculations.
Tips
- Forgetting to use combinations: Some might incorrectly think to use permutations instead of combinations, which count ordered selections.
- Improper factorial calculations: Ensure ( n! ) is correctly calculated, particularly for large ( n ).
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