In mathematics, a combination refers to the selection of items from a larger set such that the order of selection does not matter. This is different from a permutation, where order is important.
For example, if you have three fruits: apple, banana, and cherry, the combinations of two fruits are {apple, banana}, {apple, cherry}, and {banana, cherry}. Notice that {apple, banana} is the same as {banana, apple} in combinations.
The number of ways to choose \( r \) objects from \( n \) distinct objects without regard to order is denoted by:
\[^nC_r \quad \text{or} \quad \binom{n}{r}\]This is read as "n choose r" and is called a combination.
The formula to calculate combinations is:
\[^nC_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}\]where \( n! \) (n factorial) is the product of all positive integers up to \( n \), and \( 0! = 1 \) by definition.
| Aspect | Permutation | Combination |
|---|---|---|
| Order | Important (AB ≠ BA) | Not important (AB = BA) |
| Formula | \( ^nP_r = \frac{n!}{(n-r)!} \) | \( ^nC_r = \frac{n!}{r!(n-r)!} \) |
| Use Case | Arrangements, rankings | Selections, groups |
The number of permutations of \( r \) objects chosen from \( n \) is related to combinations by:
\[^nP_r = ^nC_r \times r!\]This is because permutations count all ordered arrangements, while combinations count unordered selections.
Combination theory is closely linked to sets and subsets. The number of subsets of size \( r \) from a set of size \( n \) is exactly \( \binom{n}{r} \). The power set of a set \( S \) is the set of all subsets of \( S \), and its cardinality is \( 2^n \) if \( n = |S| \).
For example, if \( S = {1, 2, 3} \), then the power set \( \mathcal{P}(S) \) has \( 2^3 = 8 \) subsets:
\[\emptyset, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}\]Pascal's Triangle (first 6 rows): 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1Note: Each number is \( \binom{n}{r} \).Combinations are fundamental in counting problems where order does not matter. They are widely used in probability, statistics, and various fields of mathematics. Understanding the difference between permutations and combinations, and mastering the formula and properties, is essential for solving entrance exam problems efficiently.
Problem: How many ways can you select 3 students from a group of 8?
Solution:
We need to find \( \binom{8}{3} \).
\[ \binom{8}{3} = \frac{8!}{3! \times 5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]Answer: 56 ways.
Problem: Find the number of permutations of 4 objects chosen from 7.
Solution:
Using the relation \( ^nP_r = ^nC_r \times r! \), first find \( \binom{7}{4} \):
\[ \binom{7}{4} = \frac{7!}{4! \times 3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]Then,
\[ ^7P_4 = 35 \times 4! = 35 \times 24 = 840 \]Answer: 840 permutations.
Problem: If \( A = { -2, -1, 0, 1, 2 } \), how many subsets of \( A \) have exactly 3 elements?
Solution:
Number of 3-element subsets is \( \binom{5}{3} \):
\[ \binom{5}{3} = \frac{5!}{3! \times 2!} = \frac{5 \times 4}{2 \times 1} = 10 \]Answer: 10 subsets.
Problem: Find the number of elements in the power set of a set with 10 elements.
Solution:
The power set has \( 2^{10} = 1024 \) elements.
Answer: 1024 subsets.
Problem: If \( n(A) = 37 \), \( n(B) = 25 \), and \( n(A \cup B) = 50 \), find \( n(A \cap B) \).
Solution:
Using the formula:
\[ n(A \cup B) = n(A) + n(B) - n(A \cap B) \]Rearranged:
\[ n(A \cap B) = n(A) + n(B) - n(A \cup B) = 37 + 25 - 50 = 12 \]Answer: 12 elements in the intersection.
Problem: If set \( A \) has 5 elements and set \( B \) has 4 elements, what is the maximum number of elements in \( A \oplus B \) (symmetric difference)?
Solution:
The symmetric difference \( A \oplus B = (A - B) \cup (B - A) \) contains elements in either \( A \) or \( B \) but not both.
The maximum occurs when \( A \cap B = \emptyset \), so no common elements.
\[ |A \oplus B| = |A| + |B| = 5 + 4 = 9 \]Answer: 9 elements.
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