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Learn the combination formula C(n,k) with examples, selection problems where order does not matter, and practice questions.
---
## Understanding Combination
Imagine you are asked to choose $$3 \text{ friends}$$ to join a futsal team. Does the order of selection matter? Of course not! What matters is **who** is selected, not the order in which they are chosen.
Visible text: Imagine you are asked to choose to join a futsal team. Does the order of selection matter? Of course not! What matters is **who** is selected, not the order in which they are chosen.
This is the fundamental difference between combination and permutation. **Combination** is a way to select a certain number of objects from a larger collection of objects, where **order is not considered**.
In daily life, we often encounter combinations when:
- Choosing food menus from a list of options
- Determining team members for an activity
- Selecting elective subjects at school
- Determining color combinations for design
The difference from permutation is very clear: if in permutation $$ABC$$ is different from $$BAC$$, then in combination $$ABC$$ is the same as $$BAC$$ because the members are the same, only the order is different.
Visible text: The difference from permutation is very clear: if in permutation is different from , then in combination is the same as because the members are the same, only the order is different.
## Combination Formula
To determine the number of ways to select $$k$$ objects from $$n$$ available objects, we use the combination formula:
Visible text: To determine the number of ways to select objects from available objects, we use the combination formula:
```math
C(n,k) = \binom{n}{k} = \frac{n!}{(n-k)! \times k!}
```
Where:
- $$C(n,k)$$ or $$\binom{n}{k}$$ means the combination of $$k$$ objects from $$n$$ objects
- $$n$$ is the total number of available objects
- $$k$$ is the number of objects to be selected
- $$n!$$ is the factorial of $$n$$
Visible text: - or means the combination of objects from objects
- is the total number of available objects
- is the number of objects to be selected
- is the factorial of
**Why is this formula different from permutation?**
The combination formula actually comes from the permutation formula divided by $$k!$$:
Visible text: The combination formula actually comes from the permutation formula divided by :
```math
C(n,k) = \frac{P(n,k)}{k!} = \frac{n!}{(n-k)! \times k!}
```
Division by $$k!$$ is done because in combinations, we **don't care about order**. Each group of $$k$$ objects has $$k!$$ different arrangement possibilities, but they are all considered **the same** in combinations.
Visible text: Division by is done because in combinations, we **don't care about order**. Each group of objects has different arrangement possibilities, but they are all considered **the same** in combinations.
## Applications in Real Situations
**Sports Team Formation:**
From $$10 \text{ available students}$$, how many ways can we select $$5 \text{ students}$$ for a basketball team?
Visible text: From , how many ways can we select for a basketball team?
Component: MathContainer
Children:
```math
C(10,5) = \frac{10!}{5! \times 5!}
```
```math
C(10,5) = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = \frac{30.240}{120} = 252 \text{ ways}
```
**Food Menu Selection:**
A restaurant offers $$8 \text{ types of food}$$ and you can choose $$3 \text{ types}$$. How many possible combination choices are there?
Visible text: A restaurant offers and you can choose . How many possible combination choices are there?
Component: MathContainer
Children:
```math
C(8,3) = \frac{8!}{5! \times 3!}
```
```math
C(8,3) = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56 \text{ combinations}
```
**Situations with Special Conditions:**
Sometimes there are **certain restrictions** in selection. For example, from $$12 \text{ students}$$ ($$7 \text{ males}$$ and $$5 \text{ females}$$), we must select $$4 \text{ students}$$ with the condition of at least $$2 \text{ females}$$.
Visible text: Sometimes there are **certain restrictions** in selection. For example, from ( and ), we must select with the condition of at least .
**Solution strategy:**
1. **Count all possibilities** that meet the conditions
2. **Separate based on conditions:** $$2 \text{ females} + 2 \text{ males}$$, $$3 \text{ females} + 1 \text{ male}$$, or $$4 \text{ females}$$
3. **Sum** all possibilities
Visible text: 1. **Count all possibilities** that meet the conditions
2. **Separate based on conditions:** , , or
3. **Sum** all possibilities
**Detailed calculations:**
**Case** $$1$$: $$2 \text{ females} + 2 \text{ males}$$
Visible text: **Case** :
Component: MathContainer
Children:
```math
C(5,2) \times C(7,2) = \frac{5!}{3! \times 2!} \times \frac{7!}{5! \times 2!}
```
```math
= \frac{5 \times 4}{2 \times 1} \times \frac{7 \times 6}{2 \times 1} = 10 \times 21 = 210
```
**Case** $$2$$: $$3 \text{ females} + 1 \text{ male}$$
Visible text: **Case** :
Component: MathContainer
Children:
```math
C(5,3) \times C(7,1) = \frac{5!}{2! \times 3!} \times \frac{7!}{6! \times 1!}
```
```math
= \frac{5 \times 4}{2 \times 1} \times 7 = 10 \times 7 = 70
```
**Case** $$3$$: $$4 \text{ females} + 0 \text{ males}$$
Visible text: **Case** :
Component: MathContainer
Children:
```math
C(5,4) = \frac{5!}{1! \times 4!} = \frac{5}{1} = 5
```
**Total:** $$210 + 70 + 5 = 285$$ ways
Visible text: **Total:** ways
## Practice Problems
1. From $$8 \text{ different books}$$, how many ways can you choose $$3 \text{ books}$$ to read during vacation?
2. A futsal team has $$12 \text{ players}$$. How many ways can they select $$5 \text{ players}$$ to play on the field?
3. In a box there are $$6 \text{ red balls}$$ and $$4 \text{ blue balls}$$. How many ways can you take $$5 \text{ balls}$$ with the condition of at least $$3 \text{ red balls}$$?
4. A student must choose $$4 \text{ subjects}$$ from $$10 \text{ available subjects}$$. If $$6 \text{ subjects}$$ are mandatory and $$4 \text{ subjects}$$ are elective, how many ways can they choose if there must be at least $$2 \text{ mandatory subjects}$$?
Visible text: 1. From , how many ways can you choose to read during vacation?
2. A futsal team has . How many ways can they select to play on the field?
3. In a box there are and . How many ways can you take with the condition of at least ?
4. A student must choose from . If are mandatory and are elective, how many ways can they choose if there must be at least ?
### Answer Key
1. **Answer: $$56 \text{ ways}$$**
Solution steps:
Given: $$n = 8$$ books, select $$k = 3$$ books
Combination formula:
```math
C(n,k) = \frac{n!}{(n-k)! \times k!}
```
```math
C(8,3) = \frac{8!}{5! \times 3!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56
```
Therefore, there are $$56 \text{ ways}$$ to choose $$3 \text{ books}$$ from $$8 \text{ available books}$$.
2. **Answer: $$792 \text{ ways}$$**
Solution steps:
Given: $$n = 12$$ players, select $$k = 5$$ players
Combination formula:
```math
C(12,5) = \frac{12!}{7! \times 5!}
```
```math
C(12,5) = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = \frac{95.040}{120} = 792
```
The futsal team can select $$5 \text{ players}$$ from $$12 \text{ players}$$ in $$792 \text{ different ways}$$.
3. **Answer: $$186 \text{ ways}$$**
Solution steps (**case method**):
Total balls: $$6 \text{ red} + 4 \text{ blue}$$ is $$10 \text{ balls}$$
Take $$5 \text{ balls}$$ with at least $$3$$ red balls
**Detailed calculation for each case:**
**Case** $$1$$: $$3 \text{ red} + 2 \text{ blue}$$
```math
C(6,3) \times C(4,2) = \frac{6!}{3! \times 3!} \times \frac{4!}{2! \times 2!}
```
```math
= \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times \frac{4 \times 3}{2 \times 1} = 20 \times 6 = 120
```
**Case** $$2$$: $$4 \text{ red} + 1 \text{ blue}$$
```math
C(6,4) \times C(4,1) = \frac{6!}{2! \times 4!} \times \frac{4!}{3! \times 1!}
```
```math
= \frac{6 \times 5}{2 \times 1} \times 4 = 15 \times 4 = 60
```
**Case** $$3$$: $$5 \text{ red} + 0 \text{ blue}$$
```math
C(6,5) \times C(4,0) = \frac{6!}{1! \times 5!} \times \frac{4!}{4! \times 0!}
```
```math
= 6 \times 1 = 6
```
**Total:** $$120 + 60 + 6 = 186$$
There are $$186 \text{ ways}$$ to take $$5 \text{ balls}$$ with at least $$3$$ red balls.
4. **Answer: $$185 \text{ ways}$$**
Solution steps (**case method**):
Mandatory subjects: $$6$$, elective subjects: $$4$$
Choose $$4 \text{ subjects}$$ with at least $$2 \text{ mandatory}$$
**Detailed calculation for each case:**
**Case** $$1$$: $$2 \text{ mandatory} + 2 \text{ elective}$$
```math
C(6,2) \times C(4,2) = \frac{6!}{4! \times 2!} \times \frac{4!}{2! \times 2!}
```
```math
= \frac{6 \times 5}{2 \times 1} \times \frac{4 \times 3}{2 \times 1} = 15 \times 6 = 90
```
**Case** $$2$$: $$3 \text{ mandatory} + 1 \text{ elective}$$
```math
C(6,3) \times C(4,1) = \frac{6!}{3! \times 3!} \times \frac{4!}{3! \times 1!}
```
```math
= \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times 4 = 20 \times 4 = 80
```
**Case** $$3$$: $$4 \text{ mandatory} + 0 \text{ elective}$$
```math
C(6,4) \times C(4,0) = \frac{6!}{2! \times 4!} \times \frac{4!}{4! \times 0!}
```
```math
= \frac{6 \times 5}{2 \times 1} \times 1 = 15 \times 1 = 15
```
**Total:** $$90 + 80 + 15 = 185$$
There are $$185 \text{ ways}$$ to choose subjects with the given conditions.
Visible text: 1. **Answer: **
Solution steps:
Given: books, select books
Combination formula:
Therefore, there are to choose from .
2. **Answer: **
Solution steps:
Given: players, select players
Combination formula:
The futsal team can select from in .
3. **Answer: **
Solution steps (**case method**):
Total balls: is
Take with at least red balls
**Detailed calculation for each case:**
**Case** :
**Case** :
**Case** :
**Total:**
There are to take with at least red balls.
4. **Answer: **
Solution steps (**case method**):
Mandatory subjects: , elective subjects:
Choose with at least
**Detailed calculation for each case:**
**Case** :
**Case** :
**Case** :
**Total:**
There are to choose subjects with the given conditions.