The concept of factorial is a prerequisite for reading this article. This concept is introduced here. For example: $4!=4\times3\times2\times 1$ (pronounced “factorial 4” not “4 factorial”).

Permutation

The number of permutations of a set is the number of “mixes” that can be made with this set.

For example: if we take a red apple (R), a blue apple (B), and a green apple (V). How many mixes can we make? We have the mixes:

• R B V
• R V B
• B V R
• V R B
• B R V
• V B R

This gives us 6 possible mixes.

Counting (the number of possibilities) can be very long! That’s why we have the formula:

$$\text{Number of permutations} = n!$$

n being the number of elements in the set.

In our example, our set $\{R,V,B\}$ has 3 elements. Using our formula, this gives us $n!=3!=3\times 2 = 6$ permutations.

Arrangement

An arrangement (without repetition) on a set is the number of possibilities of taking $k$ elements from a set of $n$ elements (taking into account the order).

Taking the previous example: if we take a red apple (R), a blue apple (B), and a green apple (V). How many ways are there (keeping the order) to take 2 apples out of these 3 apples?

• A B
• B A
• A C
• C A
• B C
• C B

This gives us 6. In the same way, there is a formula to calculate the number of arrangements easily. Here it is:

$$A^k_n=\frac{n!}{(n-k)!}$$

In our case $k = 2$ and $n=3$. So we have $\frac{3!}{(3-2)!}$ which gives us $\frac{6}{1}=6$ possibilities.

To develop your intuition, $A^3_3 = 3!$. And this is true for all $n=k$. The arrangement is an “extension” of the number of permutations of a set, we are just looking to count the number of ordered parts of this set.

Combinations

The number of combinations of a set is the number of possibilities of having $k$ elements among $n$ elements (without taking into account the order, without repetition).

Let’s take the same example: if we take a red apple (R), a blue apple (B), and a green apple (V). How many ways are there (without order) to take 2 apples out of these 3 apples?

• A B
• A C
• B C

For a set of 3 elements, we thus have 3 combinations.

The formula is once again very similar:

$$C^k_n=\binom{n}{k}=\frac{n!}{k!(n-k)!}$$

In our example: $C^2_3=\binom{3}{2}=\frac{3!}{2!(3-2)!} = \frac{6}{2} = 3$

To learn more about combinations (with exercise): Binomial Coefficient - k among n

Arrangement with Repetitions

The arrangement with repetitions is similar to the arrangement without repetition but is not calculated in the same way.

Let’s take a different example: How many possibilities are there in a 3-digit decimal padlock?

There are:

• 0 0 0
• 0 0 1
• 0 0 2
• …
• 9 9 9

Which makes 1000 possibilities.

Let’s take another example: How many possibilities are there in a 3-digit binary padlock?

We have:

• 0 0 0
• 0 0 1
• 0 1 0
• 0 1 1
• 1 0 0
• 1 0 1
• 1 1 0
• 1 1 1

Which makes 8 possibilities.

The formula to count the number of arrangements with repetitions is as follows:

$$n^k$$

In our first example, we have $n=10$ and $k=3$ which gives us $10^3=10\times10\times 10 =1000$.

In our second example, we have $n=2$ and $k=3$ which gives us $2^3=2\times 2\times2 = 8$

Summary Table (with examples)

Comment

Of course, counting is not limited to these 4 methods; there are a multitude of ways to count. If you want to see another method, we propose a solved exercise here.