While working with algorithm, you may have to calculate on the fly the number of possibilities created by some subset of data. Particularly when debugging with your \(r\) elements from \(n\) dynamic objects. Let’s take a refresh our memories!

Permutations and combination are part of the Combinatorics which is an area of mathematics about counting. A permutation is an arrangement of selected ordered items, while a combination only concerns the selection of item.

Basics refresher

Let’s review the fundamental, just to be sure we are on the same line when reviewing the following formulas. See math can be useful, to do … more math, ahem, let’s get started. 🙉


The factorial function is noted \(n!\) which represent the multiplication of all whole numbers in \(\N\) starting from \(1, 2, 3, ... n\).

For example, factorial of 4 would be calculated such as:

\[\begin{equation} \begin{split} 4! &= 4 \times 3 \times 2 \times 1 \\ &= 24 \nonumber \end{split} \end{equation}\]

The special case with \(0! = 1\) 😲:

  • You can write factorial such as \(n! = n \times (n-1)!\)
  • Meaning \(1! = 1 \times 0!\) so it means that \(0!=1\) to stay consistent.


There’s also the exponentiation, sometimes called power function which is noted \(b^n\) for \(b\) raised to the power of \(n\).

For positive numbers, when \(n\) is in \(\N\), we have \(b^n\) corresponding to multiplication of \(n\) times the base \(b\). For example \(2\) raised to the power of \(4\) would be calculated such as:

\[\begin{equation} \begin{split} 2^4 &= 2 \times 2 \times 2 \times 2 \\ &= 16 \nonumber \end{split} \end{equation}\]

Notes about exponents:

  • Noticeable cases for one; \(b^1 = b\) and zero; \(b^0 = 1\)
  • Using negative exponents; \(b^{-1} = \frac{1}{b}\)
  • Arithmetic operations; \(b^x + b^y = b^{x+y}\)


Now that we have our bases refreshed to the power of knowledge, we can factor it to our next three topics! Let’s “Q.E.D.” (_quod erat demonstrandum) those formulas and feel smart about it. 🤓



When the order does matter you want to calculate the possible permutations. Let’s say you have a set of \(n\) objects, and you choose \(k\) times from it.

\[\overbrace{(\textcolor{#ea008e}{A}, \textcolor{#007cdb}{B}, \textcolor{#00d9c2}{C})}^{\text{n=3}} \Rightarrow \underbrace{(\textcolor{#ea008e}{A}, \textcolor{#007cdb}{B}), \overbrace{(\textcolor{#ea008e}{A}, \textcolor{#00d9c2}{C})}^{\text{k=2}}, (\textcolor{#007cdb}{B}, \textcolor{#ea008e}{A}), (\textcolor{#007cdb}{B}, \textcolor{#00d9c2}{C}), (\textcolor{#00d9c2}{C}, \textcolor{#ea008e}{A}), (\textcolor{#00d9c2}{C}, \textcolor{#007cdb}{B})}_{\text{6 permutations}}\]

There is \(n\), then \(n-1\), … items to choose from, until you have \(k\) items selected. In the end you end up with \(n-k\) items that are not selected.


This looks like a factorial, so choosing \(k\) different items out of the \(n\) possible, is the same as having the \(k\) factorial from \(n\) to \(n-k\). Which can be written as:

\[\begin{equation} \begin{split} n! &= n \times n-1 \times ... \times n-(k-1) \times n-k \times ... \times 1 \\ &= \underbrace{n \times n-1 \times ... \times n-(k-1)}_{\text{number of permutations}} \times (n-k)! \nonumber \end{split} \end{equation}\]

Let’s isolate the amount of permutations on one side of the equation 🤔
dividing by \((n-k)!\) to get:

\[\begin{equation} \begin{split} \frac{n!}{(n-k)!} &= \frac{n \times n-1 \times ... \times n-(k-1) \times (n-k)!}{(n-k)!} \\ &= \frac{n \times n-1 \times ... \times n-(k-1) \times \cancel{(n-k)!}}{ \cancel{(n-k)!}} \\ &= \underbrace{n \times n-1 \times ... \times n-(k-1)}_{\text{number of permutations}} \nonumber \end{split} \end{equation}\]

So now we can write the formula to find the amount of permutation \(P_k^n\) for \(k\) selection in \(n\) objects such as:


We can see that in the case where \(k=n\) we have a division by \(0!\) and \(P_n^n=n!\)

For an example where \(n=5\) and \(k=3\), we can use it such as \(\frac{5!}{(5-3)!}=60\). With repetition (picking the same object more than once in your set), you have \(n^k\) possible permutations.



When the order does not matter you want to calculate possible combinations. There will be fewer combinations possible for a given \(k\) selections in \(n\) items than the permutation.

Let’s have an example for \(n,k=3\), we’d have one combination but six permutations:

\[\underbrace{(\textcolor{#ea008e}{A}, \textcolor{#007cdb}{B}, \textcolor{#00d9c2}{C})}_{\text{1 combination}} \Leftrightarrow \underbrace{(\textcolor{#ea008e}{A}, \textcolor{#007cdb}{B}, \textcolor{#00d9c2}{C}) (\textcolor{#007cdb}{B}, \textcolor{#00d9c2}{C}, \textcolor{#ea008e}{A}), (\textcolor{#00d9c2}{C}, \textcolor{#ea008e}{A}, \textcolor{#007cdb}{B}), (\textcolor{#ea008e}{A}, \textcolor{#00d9c2}{C}, \textcolor{#007cdb}{B}), (\textcolor{#00d9c2}{C}, \textcolor{#007cdb}{B}, \textcolor{#ea008e}{A}), (\textcolor{#007cdb}{B}, \textcolor{#ea008e}{A}, \textcolor{#00d9c2}{C})}_{\text{6 permutations}}\]

The combination is usually noted \(\binom{n}{k}\) or \(C_k^n\).


To remember the formula, there’s a little trick. In our example for \(k=n\) there are 6 times more permutations than combination. This corresponds to the number of possible arrangement, i.e. the factorial of the number of selection.

Which can be extrapolated into \(k!\) more permutations than combinations.
So the combination formula can be written such as:


Combinations’ properties:

  • Special value of \(k\) with \(C_0^n=C_n^n=1\) and \(C_1^n=C_{n-1}^n=n\)
  • Pascal’s rule: \(C_k^n+C_{k-1}^n=C_k^{n+1}\)

For repetitions, it means you can pick the same item three times like in \((\textcolor{#ea008e}{A}, \textcolor{#ea008e}{A}, \textcolor{#ea008e}{A})\) or two times like in \((\textcolor{#007cdb}{B}, \textcolor{#007cdb}{B}, \textcolor{#00d9c2}{C})\).

To find the amount of combination possible with repetition, there are multiple tricks to remember it:

  • One way: It’s to count the amount of time you pick an item \(k\) and the amount of time you skip an item \(n-1\) (because at least one item must be picked which will be used for all \(k\) selections).
  • Another way: It’s to sum possible items \(n\) with the possible amount of time they can be repeated in the selection \(k-1\).

Meaning it’s like doing the combination for \(\binom{k+n-1}{k}=\frac{(k+n-1)!}{k!(n-1)!}\).



Let’s figure out the possibilities from these three sets:

\[\overbrace{(A1, A2)}^{\text{\textcolor{#ea008e}{A}}}, \overbrace{(B1, B2, B3)}^{\text{\textcolor{#007cdb}{B}}}, \overbrace{(C1, C2, C3)}^{\text{\textcolor{#00d9c2}{C}}}\]

The number of element of one set is also called cardinalilty noted as \(|A|\) or \(card(A)\), they differ between each set. Meaning we have 3 choice and each one have respectively 2, 3, 3 possible options such as each possibility can be noted as a set of one element from \(\textcolor{#ea008e}{A}\), \(\textcolor{#007cdb}{B}\) and \(\textcolor{#00d9c2}{C}\) such as:

\[\underbrace{(A2, B1, C3)}_{\text{one possible arrangement}}\]

In this case there’s no need to try to apply the previous permutation or combination formula.


The simplest way to calculate the possibility between asymmetric subsets is called the rule of product. I like the Cartesian product notation:

\[\{ A\times B = \{ (a,b)~|~a \in A, b \in B \}\]

The product of two set \(A\) and \(B\) is another set composed of pairs \((a, b)\) where \(a\) is an element of \(A\) and \(b\) one of \(B\). So for our three disjointed sets presented earlier, there’s \(|\textcolor{#ea008e}{A}| \times |\textcolor{#007cdb}{B}| \times |\textcolor{#00d9c2}{C}|\) possible permutations, in our case it amounts to \(18\).

Words of wisdom

That’s it! Let’s use our new-found knowledge soon when working with multiple datasets, arrays or any other type of weird data structures.

As a side note, I would prefer saying combinatronics 🤖 instead of combinatorics, it sounds way better in my opinion. Feeling like we’re in some sci-fi 80’s scientific cliché.