How many subsets of cardinality 3 does A have if |A| = n?
Understand the Problem
The question is asking how many subsets of cardinality 3 can be formed from a set A, given that the size of set A is n. This involves combinatorial mathematics, particularly the concept of combinations.
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 the set ( A ) is given by: $$ \frac{n(n-1)(n-2)}{6} $$
Steps to Solve
- Understanding the Combinations Formula
To find the number of subsets of cardinality 3 from a set ( A ) of size ( n ), we use the combinations formula: $$ C(n, k) = \frac{n!}{k!(n-k)!} $$ where ( k ) is the number of elements to choose (in this case, 3).
- Substituting Values into the Formula
For our specific case, substitute ( k = 3 ): $$ C(n, 3) = \frac{n!}{3!(n-3)!} $$
- Simplifying the Combinations Expression
The factorials simplify as follows:
- ( 3! = 6 )
- The expression becomes: $$ C(n, 3) = \frac{n(n-1)(n-2)}{6} $$
- Final Formula for Cardinality 3 Subsets
Thus, the number of subsets of cardinality 3 from the set ( A ) is: $$ \frac{n(n-1)(n-2)}{6} $$
The number of subsets of cardinality 3 from the set ( A ) is given by: $$ \frac{n(n-1)(n-2)}{6} $$
More Information
This formula calculates how many different ways you can select 3 elements from a total of ( n ) elements. This is an important concept in combinatorics, often used in probability and statistics.
Tips
- Forgetting that ( n ) must be at least 3 for subsets of cardinality 3 to exist. If ( n < 3 ), ( C(n, 3) ) equals zero.
- Miscalculating factorials or the combinations formula. Always double-check the simplifications.
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