# How many possible matchups are there in League of Legends?

How many possible matchups are there in *League of Legends*? The answer depends on a couple things, including the number of champions in the game and what you mean by “matchup” - there are a lot more blind pick matchups than there are ranked matchups. In this post, we’ll answer both of these questions and give you all the combinitorial tools you need to answer these and many more such questions!

If you’re already familiar with combinatorics (or don’t care), you can skip to the answer.

## Defining factorial: What is a combination?

### Permutations of `n`

elements

First, let’s answer an easier question. How many possible team combinations are there in a game with 2 choices for champion (A, B) and teams of size 2? Let’s say for now that order DOES matter.

It’s not a trick question, there’s two:

- A B
- B A

If there are \(3\) champions and teams of size 3 (and order still matters), then we have \(6\) possibilities:

- A B C
- A C B
- B A C
- B C A
- C A B
- C B A

Let’s look a bit closer at this. Notice how we have \(3\) options for the first item (2 `A`

s then 2 `B`

s then 2 `C`

s), and then after that it looks a lot like repetitions of the answer for 2 teams? Or if we look a bit differently, there’s **three** choices for the first slot, then for each first champion there are **two** choices for the second slot, and finally for each set of first 2 champions, there’s **one** choice for the last slot. “For each” means we should multiply these numbers together, and we get: \(3 \cdot 2 \cdot 1 = 6\).

Does this hold up for 4 teams? Well, let’s see. If we have A, B, C, D then:

- A can be first, and then we have 6 options for B, C, D
- B can be first, and then we have 6 options for A, C, D
- C can be first, and then we have 6 options for A, B, D
- D can be first, and then we have 6 options for A, B, C

\(6 + 6 + 6 + 6 = 4 \cdot 6 = 24\). In other words, the total number of possible **permutations** of \(4\) elements is \(4 \cdot 3 \cdot 2 \cdot 1 = 24\).

In math, we have a special symbol for this operation, called **factorial** and written with an \(!\):
$$5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120$$
As you might guess, factorial grows pretty quickly.

But the number of possible teams in League of Legends with \(N\) total champions is a LOT less than \(N!\) - after all, the team sizes are only \(5\), not \(N\). And, order doesn’t matter. Let’s look at a few more examples. First, we’ll address what to do when we have more champions than team slots.

### Permutations of `r`

elements, `r < n`

We’ll start with \(3\) champions, and look at team comps for teams of size \(2\) and \(1\).

With teams of size \(2\), we can do:

- B C (A is missing)
- C B (A is missing)
- A C (B is missing)
- C A (B is missing)
- B A (C is missing)
- A B (C is missing)

It’s like we didn’t even get rid of anything! Well, this makes sense. If we are sorting \(3\) champions, we’re sorting between Top, Mid, and Bot. And if we’re sorting only \(2\) champions, let’s say Top and Bot, it turns out that we end up “sorting” our final champion into a slot – a pretend slot containing champions that aren’t part of our team! Let’s call that pretend-slot “Missing.” Either way, we have \(3\) champions, and \(3\) slots; either we choose one for Mid and then sort the other two into Top and Bot, or we choose one for Missing (not picked) and then sort the other two into Top and Bot.

With teams of size \(1\), here are the options:

- C (A and B are missing)
- B (A and C are missing)
- A (C and B are missing)

If we’re only picking a *single* champion (say it’s a top laner, top lane is an island so top is the only player on the team), though, we put \(2\) champions into Missing and then sort one into the remaining lane. But *this* time, we’re changing the outcome, because “A and B are missing” is the same as “B and A are missing.” So for each champion that we srt into top lane, \(2\) of our initial \(6\) orderings are accounted for, because we don’t count the order of the champions that are Missing.

Let’s look at \(4\) champions to make this a little clearer:

With teams of size \(4\), we know there will be \(24\) possible orderings (\(4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24\)). And with teams of size \(3\), there will *also* be \(24\) possible orderings (Top, Mid, Bot, and Missing)

But what about teams of size \(2\)?

- C D (A and B are missing)
- D C (A and B are missing)
- B D (A and C are missing)
- B D (A and C are missing)
- B C (A and D are missing)
- C B (A and D are missing)
- A D (B and C are missing)
- D A (B and C are missing)
- A C (B and D are missing)
- C A (B and D are missing)
- A B (C and D are missing)
- B A (C and D are missing)

And if teams have size \(1\), then there’s four:

- D (A, B, C are missing)
- C (A, B, D are missing)
- B (A, C, D are missing)
- A (B, C, D are missing)

So, if there are \(2\) champions on a team, then once again each possibility represents \(2\) different orderings of our total \(4\) champions (`C D`

represents `C D A B`

and `C D B A`

; `D C`

represents `D C A B`

and `D C B A`

; etc.), so the answer is \(\frac{4!}2\) or \(\frac{24}2=12\).

And if there is \(1\) champion on a team, then each possibility represents \(6\) different orderings out of our total \(24\), or all the ways to order \(3\) champions (`D`

represents `A B C`

, `A C B`

, `B A C`

, `B C A`

, `C A B`

, and `C B A`

).

In other words, the answer is \(\frac{24}6\). But that’s really \(\frac{24}{3!}\), or \(\frac{4!}{3!}\). And in fact, the \(12\) ways to order \(2\) champions isn’t \(\frac{4!}2\), it’s \(\frac{4!}{2!}\). Cool!

In math, we also have a term for this formula. We say that “4 permute 2” is \(\frac{4!}{(4 - 2)!}\), and “4 permute 1” is \(\frac{4!}{(4-1)!}\). On a calculator, or in Wolfram|Alpha, you can use the notation `nPr(4, 2)`

or `nPr(4, 1)`

.

In a general form, “n permute r” (the number of ways to permute \(r\) items out of \(n\) options) is: $$ \frac{n!}{(n-r)!} $$

### Combinations of `r`

elements, `r < n`

To conclude our whirlwind tour of counting problems, let’s take a look at **combinations** rather than **permutations**, i.e. team comps where order doesn’t matter.

If we have \(n\) champions and our team size (represented by \(r\)) is also \(n\), i.e. \(r=n\), there is of course only \(1\) team comp possible (no repeated champions), the one that uses every available champion one time. But what if the team size is smaller than \(n\), i.e. \(r < n\)?

Let’s say there are \(3\) champions, and we want to pick \(2\) of them. Then here are the options:

- B C (A is missing)
- A C (B is missing)
- B A (C is missing)

And now let’s look at the options if we want to pick 1:

- A (B and C are missing)
- B (A and C are missing)
- C (A and B are missing)

Hmm, that looks a little familiar, doesn’t it?

Let’s look at the situation with \(4\) champions in our pool also.

If the team size is \(3\), then we have \(4\) options:

- B C D (A is missing)
- A C D (B is missing)
- A B D (C is missing)
- A B C (D is missing)

If it’s \(2\), then we have \(6\) options:

- A B (C and D are missing)
- A C (B and D are missing)
- A D (B and C are missing)
- B C (A and D are missing)
- B D (A and C are missing)
- C D (A and B are missing)

And if it’s \(1\), then there are \(4\) options:

- A (B, C, and D are missing)
- B (A, C, and D are missing)
- C (A, B, and D are missing)
- D (A, B, and C are missing)

Notice, the cases of team size \(1\) and team size \(3=4-1\) look *very* similar to each other. In fact, they are the same case! Because when order doesn’t matter, we have 2 bins: “part of the team” and “not part of the team.” Order doesn’t matter in *either* of these bins. So, picking 3 and not picking 1 is the same as picking 1 and not picking 3.

If our teams are size \(r\) and there are \(n\) champions total, then there are $$\frac{n!}{(n-r)! \cdot r!}$$ total possible team comps (where order doesn’t matter). This is called “n choose r,” and on a calculator or Wolfram|Alpha it can be written `ncr(4, 1)`

. In math notation, it’s often written like this: $$\binom{n}{r}$$ or for a specific example of “n choose 1”: $$\binom{4}{1}$$

## The number of unique team comps in League of Legends

Since the number of champions in *League of Legends* is constantly changing, we’ll give you general answers in terms of \(n\), where \(n\) is the number of champions in the game. You can then evaluate them on Wolfram|Alpha by changing the formula slightly.

### The number of unique single teams

If we only care about one team, and not a full matchup, then there are `nCr(n, 5)`

total possible teams. We can evaluate this at n=167 and find there are \({1,018,963,693}\) total possible team comps (order doesn’t matter, and just a single team)

### The number of matchups in ranked

In ranked, we can’t repeat champions, and order doesn’t matter. So instead of 2 buckets (“chosen” and “not chosen”), we have 3 buckets (“team 1,” “team 2,” and “not chosen”). Furthermore, “team 1” and “team 2” *themselves* can be switched around without a new matchup occurring. So, the answer is:
$$
\frac{n!}{5! \cdot 5! \cdot (n - 10)! \cdot 2!}
$$Evaluating this at n=167 gives us \(445,105,081,711,753,848\).

Alternatively, we can say that we have `nCr(n, 5)`

combinations for the first team, and then for each of those combinations, there are `nCr(n-5, 5)`

combinations for the second team; and then divide by 2 because Team 1 vs Team 2 is the same as Team 2 vs Team 1. Remember, “for each” means “multiply,” so our formula is:

$$ \frac{\text{nCr}(n, 5) \cdot \text{nCr}(n - 5, 5)}{2!} $$ $$ =\frac{n! \cdot (n-5)!}{(n-5)! \cdot 5! \cdot (n-5-5)! \cdot 5! \cdot 2!} $$ Reducing this, we get: $$ \frac{n!}{(n-5-5)! \cdot 5! \cdot 5! \cdot 2!} $$ $$ =\frac{n!}{5! \cdot 5! \cdot (n-10)! \cdot 2!} $$ which is the same as our previous formula!

### The number of matchups in blind pick

In blind pick, it’s a bit more complicated. In ranked, since we encounter `A B C D E vs F G H I J`

**and** `F G H I J vs A B C D E`

, we need to divide by \(2!\) at the end. However, in blind pick, we can encounter `A B C D E vs A B C D E`

(a full mirror matchup), but we’ll only encounter it a single time! Uh oh!

If we didn’t have to worry about this discrepancy in the amount of double-counting, then the answer would be simply:

$$
\frac{(\text{nCr}(n, 5))^2}{2!}
$$
We build two teams independently with `nCr(n, 5)`

and then divide by \(2!\) to make team 1 and team 2 interchangeable. So let’s start here and then figure out how many matchups we’ve now missed by dividing by \(2!\). For each team 1, there is exactly one team 2 that would make for a mirror matchup, so the number of mirror matchups is the same as the number of individual team comps, or `nCr(n, 5)`

. And right now, we’ve undercounted these by a factor of two, since we divided *everything* by 2 but we should not have divided the mirror matchups by anything.

So, our final answer is $$ \frac{(\text{nCr}(n, 5))^2}{2!} + \frac{\text{nCr}(n, 5)}{2!} $$ or $$ \frac{\text{nCr}(n, 5)^2 + \text{nCr}(n, 5)}{2!} $$ Let’s once again evaluate this at n=167, and we get our answer: \(519,143,504,335,580,971\). For reference, the universe began about \(436,117,076,600,000,000\) seconds ago.

Naturally, you’re probably now wondering how many champions *League of Legends* would need to have in order to have as many blind-pick team comps as there are atoms in the universe. The number of atoms in the universe isn’t known exactly, but it’s approximately \(10^{80}\).
If *League* ever gets to 600,000 total champions, then we’ll reach 136042093708252960986105099288464763802542437861477225410796760000864000000000000 or \(1.3 \times 10^{80}\) total possible blind-pick team comps!