Combination

Understanding Combination

Imagine you are asked to choose 3 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.

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.

Combination Formula

To determine the number of ways to select $k$ objects from $n$ available objects, we use the combination formula:

Where:

• $C(n,k)$ or $\binom{n}{k}$ = combination of $k$ objects from $n$ objects
• $n$ = total available objects
• $k$ = number of objects to be selected
• $n!$ = factorial of $n$

Why is this formula different from permutation?

The combination formula actually comes from the permutation formula divided by $k!$:

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.

Applications in Real Situations

Sports Team Formation:

From 10 available students, how many ways can we select 5 students for a basketball team?

Food Menu Selection:

A restaurant offers 8 types of food and you can choose 3 types. How many possible combination choices are there?

Situations with Special Conditions:

Sometimes there are certain restrictions in selection. For example, from 12 students (7 males and 5 females), we must select 4 students with the condition of at least 2 females.

Solution strategy:

1. Count all possibilities that meet the conditions
2. Separate based on conditions: 2 females + 2 males, 3 females + 1 male, or 4 females
3. Sum all possibilities

Detailed calculations:

Case 1: 2 females + 2 males

Case 2: 3 females + 1 male

Case 3: 4 females + 0 males

Total: $210 + 70 + 5 = 285$ ways

Practice Problems

1. From 8 different books, how many ways can you choose 3 books to read during vacation?

2. A futsal team has 12 players. How many ways can they select 5 players to play on the field?

3. In a box there are 6 red balls and 4 blue balls. How many ways can you take 5 balls with the condition of at least 3 red balls?

4. A student must choose 4 subjects from 10 available subjects. If 6 subjects are mandatory and 4 subjects are elective, how many ways can they choose if there must be at least 2 mandatory subjects?

Answer Key

1. Answer: 56 ways

   Solution steps:

   Given: $n = 8$ books, select $k = 3$ books

   Combination formula:

   $$C(n,k) = \frac{n!}{(n-k)! \times k!}$$

   $$C(8,3) = \frac{8!}{5! \times 3!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56$$

   Therefore, there are 56 ways to choose 3 books from 8 available books.

2. Answer: 792 ways

   Solution steps:

   Given: $n = 12$ players, select $k = 5$ players

   Combination formula:

   $$C(12,5) = \frac{12!}{7! \times 5!}$$

   $$C(12,5) = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = \frac{95.040}{120} = 792$$

   The futsal team can select 5 players from 12 players in 792 different ways.

3. Answer: 186 ways

   Solution steps (case method):

   Total balls: 6 red + 4 blue = 10 balls

   Take 5 balls with at least 3 red balls

   Detailed calculation for each case:

   Case 1: 3 red + 2 blue

   $$C(6,3) \times C(4,2) = \frac{6!}{3! \times 3!} \times \frac{4!}{2! \times 2!}$$

   $$= \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times \frac{4 \times 3}{2 \times 1} = 20 \times 6 = 120$$

   Case 2: 4 red + 1 blue

   $$C(6,4) \times C(4,1) = \frac{6!}{2! \times 4!} \times \frac{4!}{3! \times 1!}$$

   $$= \frac{6 \times 5}{2 \times 1} \times 4 = 15 \times 4 = 60$$

   Case 3: 5 red + 0 blue

   $$C(6,5) \times C(4,0) = \frac{6!}{1! \times 5!} \times \frac{4!}{4! \times 0!}$$

   $$= 6 \times 1 = 6$$

   Total: $120 + 60 + 6 = 186$

   There are 186 ways to take 5 balls with at least 3 red balls.

4. Answer: 185 ways

   Solution steps (case method):

   Mandatory subjects: 6, elective subjects: 4

   Choose 4 subjects with at least 2 mandatory

   Detailed calculation for each case:

   Case 1: 2 mandatory + 2 elective

   $$C(6,2) \times C(4,2) = \frac{6!}{4! \times 2!} \times \frac{4!}{2! \times 2!}$$

   $$= \frac{6 \times 5}{2 \times 1} \times \frac{4 \times 3}{2 \times 1} = 15 \times 6 = 90$$

   Case 2: 3 mandatory + 1 elective

   $$C(6,3) \times C(4,1) = \frac{6!}{3! \times 3!} \times \frac{4!}{3! \times 1!}$$

   $$= \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times 4 = 20 \times 4 = 80$$

   Case 3: 4 mandatory + 0 elective

   $$C(6,4) \times C(4,0) = \frac{6!}{2! \times 4!} \times \frac{4!}{4! \times 0!}$$

   $$= \frac{6 \times 5}{2 \times 1} \times 1 = 15 \times 1 = 15$$

   Total: $90 + 80 + 15 = 185$

   There are 185 ways to choose subjects with the given conditions.