Dear Uncle Colin,

I’m trying to figure out how many possible draws there are for the Champions’ League quarter-finals. There are eight teams involved, and let’s assume first leg home advantage doesn’t matter (so A vs B is the same as B vs A) and there’s no restriction on which teams can be matched. How would you work it out?

- Understanding Every Football Algorithm

Hi, UEFA, and thanks for your message!

Here’s how I would work it out. I would:

- Arrange the teams into some sort of order (e.g., alphabetical).
- Notice that the first team in the list has seven possible opponents.
- Look at the remaining six teams: the first team on that list has five possible opponents.
- Etc. and so on.

All in all, there are $7 \times 5 \times 3 \times 1 = 105$ possible draws.

### But what if home advantage is important?

In that case, each of the 105 distinct draws could be arranged in any one of 16 ways (each of the four matches could be reversed).

That gives a total of $105 \times 16 = 1680$ possible draws.

Hope that helps!

- Uncle Colin

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.