Combination

Selections in Which Order Does Not Matter

Suppose you choose $3 \text{ friends}$ for a futsal team. Selecting Ana, Budi, and Citra gives the same team in any order. What matters is who is selected.

That is the central difference between combinations and permutations. A combination selects objects from a larger set when order does not matter.

In daily life, we often encounter combinations when:

• Choosing dishes from a menu
• Determining team members for an activity
• Selecting elective subjects at school
• Determining color combinations for design

For a permutation, $ABC$ and $BAC$ are different arrangements. For a combination, they represent the same selection because their members are identical.

Combination Formula

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

The variables in this formula are:

• $C(n,k)$ 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 follows by dividing the permutation count by $k!$:

We divide by $k!$ because a selection of $k$ objects can be arranged in $k!$ orders. A combination counts all of those orders as one selection.

Counting Selections With and Without Conditions

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 $8$ dishes and you can choose $3$. How many selections are possible?

Situations with Special Conditions:

Some selections include additional conditions. For example, from $12 \text{ students}$ ($7 \text{ males}$ and $5 \text{ females}$), select $4 \text{ students}$ with 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

Try each problem before opening the worked solution. First decide whether order matters, then list every valid case without overlap.

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 chooses $4 \text{ subjects}$ from $10 \text{ available subjects}$. If $6 \text{ subjects}$ are sciences and $4 \text{ subjects}$ are humanities, how many selections contain at least $2 \text{ science subjects}$?

Worked Solutions

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

   Solution steps:

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

   Combination formula:

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

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

   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:

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

   $$\begin{aligned} C(12,5) &= \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} \\ &= \frac{95\,040}{120} \\ &= 792 \end{aligned}$$

   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: $3 \text{ red} + 2 \text{ blue}$

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

   $$\begin{aligned} &= \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times \frac{4 \times 3}{2 \times 1} \\ &= 20 \times 6 \\ &= 120 \end{aligned}$$

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

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

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

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

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

   $$= 6 \times 1 = 6$$

   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):

   Science subjects: $6$, humanities subjects: $4$

   Choose $4 \text{ subjects}$ with at least $2 \text{ sciences}$

   Detailed calculation for each case:

   Case 1: $2 \text{ sciences} + 2 \text{ humanities}$

   $$\begin{aligned} &C(6,2) \times C(4,2) \\ &= \frac{6!}{4! \times 2!} \times \frac{4!}{2! \times 2!} \end{aligned}$$

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

   Case 2: $3 \text{ sciences} + 1 \text{ humanity}$

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

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

   Case 3: $4 \text{ sciences} + 0 \text{ humanities}$

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

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

   Total: $90 + 80 + 15 = 185$

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