Apakah kombinasi itu?

Let's look at an example first. Suppose there are $7$ people and we have to form a committee of $3$ people out of these $7$. (In the committee all $3$ have equal power!)

Let's think about how many ways we can do this.

For the time being, let us assume that the committee's members are represented as $M_1,M_2$ and $M_3$. The $7$ people are denoted as $P_1,P_2,\ldots,P_7$. Then, as explained under What is a permutation?, the number of ways of forming the committee is $7\times6\times5=210$. But we have to notice one thing! Here, we have counted every committee that satisfies our needs six times.

We have considered the committee formed by $P_2,P_4$ and $P_7$ six times. That's because we have considered the following committees to be distinct:

1. $M_1=P_2,M_2=P_4,M_3=P_7$

2. $M_1=P_2,M_2=P_7,M_3=P_4$

3. $M_1=P_4,M_2=P_2,M_3=P_7$

4. $M_1=P_4,M_2=P_7,M_3=P_2$

5. $M_1=P_7,M_2=P_2,M_3=P_4$

6. $M_1=P_7,M_2=P_4,M_3=P_2$

Whereas for our purposes, these should not be considered distinct, because as we said, in the committee every one has equal power — so they cannot be labelled as $M_1,M_2$ and $M_3$. (Because we may consider $M_1$ as the chairperson and give $M_1$ more power than the others!)

We should divide $210$ by $6$. (As then we will be considering one committee exactly $6/6=1$ time).

So, our answer is $210/6=35$.

(Note that $6=3!$, i.e., the number of ways that $3$ distinguishable objects can be arranged among themselves.)

We say that there are $35$ combinations of $3$ people from $7$ people.

So, a combination is a permutation where the order of the objects is not important.

Since $k$ distinguishable objects can be arranged among themselves in $k!$ distinct ways, so if there are $p$ permutations of $k$ objects from $n$ distinguishable objects, then how many combinations of $k$ objects from the $n$ distinguishable objects are there?

Yes, the required number is $\frac{p}{k!}$.

The number of combinations of $r$ objects out of a set of $n$ is given by the formula $$\binom{n}{r} = \frac{n!}{r!(n-r)!}$$ You might also see combinations notated in several different ways. All of these have the exact same meaning. $$\binom{n}{r} = {}_{n} C_r = C(n,r)$$ Now you might be looking at the abundance of factorials and wondering how the equation even makes sense, so let's walk through it step by step!

Let's say we have $10$ tennis balls and we want to find the number of ways to choose a combination of $3$ of them.

Let's choose the balls one at a time. For the first ball, we have $10$ balls to choose from. For the second, we have $9$ remaining balls to choose from, and logically following, we have $8$ choices for the third ball. This means that there are a total of $10\cdot 9\cdot 8$ ways to choose the three balls.

Now, here's the catch — the balls we chose aren't distinguishable! If we had switched the order in which we chose the three balls we now have, the end result would still be the same: a group of $3$ balls in our hands, ready to be played.

The number of ways these three balls could be ordered is $3\cdot 2\cdot 1$. This means that overall, the number of ways we can pick $3$ indistinguishable balls from a group of $10$ is $$\frac{10\cdot 9\cdot 8}{3\cdot 2\cdot 1} = \boxed{120}.$$

Let's think about how this works with all the factorials in our combination formula. If we multiply the numerator and denominator by $7!$, we find the following: $$\frac{10\cdot 9\cdot 8\cdot 7!}{3\cdot 2\cdot 1\cdot 7!} = \frac{10!}{3!\cdot 7!} = \frac{10!}{3!\cdot (10-3)!} = \boxed{\binom{10}{3}}$$ which works out with our formula! Now all the factorials make sense.

Here's a quick bit of practice:

For more fun math: a closely related concept to combinations is permutations. The only difference is that order does matter — this difference is reflected in a very simple change in our equation.

(Video) Combinations made easy

by tecmath

Check out this video by tecmath to learn more about combinations!

##Combinations

According to the video, combinations are specifically finding probability when order doesn't matter. Let's dive deeper.

The example used in the video is that there are a total of 5 books, and we want to select 3 books.

For our first choice, we have 5 options to choose from. For the next book, we have 4 options to chose from. For the last book, we have 3 options to choose from.

Therefore we can multiply: $5 \times 4 \times 3 = 60$. You may notice that this is the same as the permutation!

But, we're saying order doesn't matter, so we need to have another step.

We know there are a total of 3 different books we can take, so to find the number of ways 3 spaces can be ordered, we can do...

$3 \times 2 \times 1$, which would give us $6$ spaces. We can also use factorials: $3!$.

Finally, to find the number of combinations, we can do...

$$\frac{5 \times 4 \times 3 }{3 \times 2 \times 1} = \frac{60}{6} = 10$$

The general formula to find the number of combinations is:

$$\frac{n!}{(n-r)!r!}$$

where...

n = number of total objects

r = number of items selected

Try the other examples in the video!

Combination Formula

In combinations, the arrangement or order of the elements selected does not matter. The order of the selection does not matter. That means that if you select the same set of items, regardless of the order in which they were selected, it's considered the same combination. Let's say we have a bag of red, blue, and green marbles. The total number of marbles in the bag is ten. Whether you pick a red then a blue marble, or you pick a blue then a red marble, the combination probability value will be the same.

This concept applies to larger sets, too. Selection order does not matter.

The formula to calculate combination is very similar to the formula for permutation.

In the realm of permutations and combinations, remember that permutation deals with arrangements where the order does matter, while combination deals with selections where the order doesn't matter.

There are types of permutation. The first is the repetition permutation, which allows repetition of elements in arrangements. Next, we have circular permutation, which deals with arrangements in a circular manner. Third, we have ordered permutation, where every arrangement is distinct. Lastly, we have unordered permutation, which is also known as combination!

Combinatorics with Pascal’s Triangle provides a visual way to find combinations. Each number in the triangle corresponds to the number of ways objects can be selected.