Combination

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.

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 $\binom{n}{k}$ 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 $n$

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 $10 \text{ available students}$, how many ways can we select $5 \text{ students}$ for a basketball team?

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?

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 $2 \text{ females}$.

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

Detailed calculations:

Case $1$: $2 \text{ females} + 2 \text{ males}$

Case $2$: $3 \text{ females} + 1 \text{ male}$

Case $3$: $4 \text{ females} + 0 \text{ males}$

Total: $210 + 70 + 5 = 285$ 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 $2 \text{ mandatory subjects}$?

Answer Key

1. Answer: $56 \text{ ways}$

   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 $56 \text{ ways}$ 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:

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

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

   The futsal team can select $5 \text{ players}$ 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 $1$: $3 \text{ red} + 2 \text{ blue}$

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

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

   Case $2$: $4 \text{ red} + 1 \text{ blue}$

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

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

   Case $3$: $5 \text{ red} + 0 \text{ blue}$

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

   $$= 6 \times 1 = 6$$

   Total: $120 + 60 + 6 = 186$

   There are $186 \text{ ways}$ 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 $1$: $2 \text{ mandatory} + 2 \text{ elective}$

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

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

   Case $2$: $3 \text{ mandatory} + 1 \text{ elective}$

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

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

   Case $3$: $4 \text{ mandatory} + 0 \text{ elective}$

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

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

   Total: $90 + 80 + 15 = 185$

   There are $185 \text{ ways}$ to choose subjects with the given conditions.