4.2 Combination
In our daily lives, we often select items or form groups where the order of selection does not matter. For instance, choosing two friends out of three to go kayaking together results in the same group whether friend A or friend B is named first. In Additional Mathematics, this process of selection without regard to order or arrangement is known as a combination.
Learning Objectives
- Compare and differentiate between permutations and combinations.
- Determine the number of combinations of \(r\) objects chosen from \(n\) distinct objects at a time using the formula \(^nC_r = \dfrac{^nP_r}{r!} = \dfrac{n!}{r!(n - r)!}\).
- Solve problems involving combinations with specific conditions, such as forming committees, grouping objects, and applying restrictions like "at least" or "not more than".
Permutation vs Combination Comparison & Formula Map
Clarify the fundamental difference between order-dependent selection and order-independent selection, while showing how \(r!\) eliminates repeated arrangements of the same group.
Permutation
Order matters. \(AB \neq BA\).
\(^nP_r = \dfrac{n!}{(n - r)!}\) Electing Chairperson & SecretaryCombination
Order does not matter. \(\{A, B\} = \{B, A\}\).
\(^nC_r = \dfrac{^nP_r}{r!} = \dfrac{n!}{r!(n - r)!}\) Choosing 2 representativesShort Explanation
Combination is a counting technique used when the sequence or order of selected items does not create a new outcome.
1. Comparing Permutation and Combination
A permutation is a process of arranging objects where the order and sequence are taken into consideration. A combination is a process of selection without considering the order and sequence of the objects.
2. Formula for the Number of Combinations
The number of combinations of \(r\) objects selected from \(n\) distinct objects is given by \(^nC_r = \dfrac{^nP_r}{r!} = \dfrac{n!}{r!(n - r)!}\).
3. Special Values and Properties
\(^nC_0 = 1\), \(^nC_1 = n\), \(^nC_n = 1\), and the symmetry property \(^nC_r = {}^nC_{n - r}\).
4. Solving Problems with Conditions
When solving combination problems with conditions, the conditions must be dealt with first. Use the multiplication rule for sequential stages and add the outcomes of mutually exclusive cases.
| Scenario | Type | Reason |
|---|---|---|
| Choosing 2 out of 5 pupils as class leader and assistant class leader | Permutation | Order matters: \(AB \neq BA\). |
| Choosing 2 out of 5 pupils to join a competition | Combination | Order does not matter: \(\{A, B\} = \{B, A\}\). |
| Selecting 3 committee members from 10 candidates | Combination | No specific posts are assigned. |
| Forming triangles from the 6 vertices of a hexagon | Combination | Any 3 vertices form the same triangle regardless of order. |
Steps for Solving Combination Problems with Conditions
Identify the conditions or restrictions in the problem.
Deal with the conditions first by selecting the required objects.
Apply the multiplication rule for sequential stages.
Add the outcomes of mutually exclusive cases.
Number of combinations of \(r\) objects from \(n\) distinct objects
\(^nC_r = \dfrac{^nP_r}{r!} = \dfrac{n!}{r!(n - r)!}\)Also written as \(\binom{n}{r}\), known as the binomial coefficient.
Try to Answer First
Answer in your mind, then press "Check Answer".
Determine whether choosing 4 out of 18 television channels involves a permutation or a combination, and find the number of possible selections.
A committee of 5 is to be elected from 5 men and 3 women. Find the number of committees that can be formed if the committee must contain exactly 3 men and 2 women.
Show that \(^nC_r = {}^nC_{n - r}\) using the combination formula.
Common Mistakes
- Using permutation (\(^nP_r\)) instead of combination (\(^nC_r\)) when selecting committee members or representatives where no specific posts or hierarchical roles are designated.
- Forgetting to divide by \(r!\) when deriving combinations from permutations.
- Forgetting to reduce the pool of remaining items when performing sequential grouping, for example choosing 3 from 8 after 2 of the original 10 have already been placed in the first group.
- Multiplying cases together instead of adding them when dealing with mutually exclusive scenarios, for example adding Case 1 and Case 2 in "at least" problems.
Concept Misunderstandings
Changing the order of choosing groups alters the final total number of ways when dividing a group of items.
Whether you select 2 people first, then 3, then 5, or select 5 people first, then 3, then 2, the product of the combinations remains identical: \(^{10}C_2 \times {}^8C_3 \times {}^5C_5 = 45 \times 56 \times 1 = 2520\); or \(^{10}C_5 \times {}^5C_3 \times {}^2C_2 = 252 \times 10 \times 1 = 2520\).
The term "combination" in everyday language, such as a "combination lock", has the same meaning in mathematics.
In mathematics, a combination strictly means order does NOT matter. A code for a lock where order matters is technically a permutation, not a mathematical combination.
Summary
- Combination Definition: Selection of objects where arrangement, order, or sequence does not matter.
- Basic Formula: \(^nC_r = \dfrac{^nP_r}{r!} = \dfrac{n!}{r!(n - r)!}\).
- Key Identities: \(^nC_0 = 1\), \(^nC_1 = n\), \(^nC_n = 1\), and \(^nC_r = {}^nC_{n - r}\).
- Problem-Solving Strategy: Deal with restrictions first, apply the multiplication rule for concurrent stages, and sum the outcomes of mutually exclusive cases.
Short Activity
Objective: Identify combinations, apply the combination formula, and match scenarios to their correct classification or value.
A. Objective Quiz
1 There are 10 candidates running for a club committee. If 3 committee members are to be selected without assigning any specific posts, how many different ways can the selection be made?
2 A football squad consists of 17 local players and 3 foreign players. In how many ways can a coach select a starting lineup of 11 players that includes exactly 2 foreign players?
B. Fill in the Blanks
3 The mathematical process of selecting objects without considering order or sequence is called a ________.
4 For any positive integer \(n\), the value of \(^nC_0\) is equal to ________.
C. Matching / Drag and Drop
Drag each item to the correct classification or value. On a phone, tap an item first, then tap the matching box.
