This is the week! The shipment which has Mice and Mystics, the newest Summoner Wars Reinforcement Packs Bellor's Retribution and Saella's Precision, and reprints of several out-of-stock items, should be arriving at Plaid Hat Games this week. After they do, several days will be needed to process all the pre-orders, and more time for things to ship to their destinations, so some patience is still required.

In the meantime, I thought I would celebrate the widespread availability of all the SWs faction reinforcements by republishing a series of articles I began at the Blagog.com blog, titled "How Many Decks?" I am using this opportunity to update the numbers to take all the final reinforcements into account. I will do one per day, and finish the series with new articles as well.

In these articles I am indulging an itch I've been wanting to scratch for a while, and combine my inner math geek and my game geek. I am investigating the question:

How many different Summoner Wars decks can you build?

I'm going to review some combinatorial math while I'm at it. You can enjoy it, or skip to the conclusions because either you know it all or your eyes have glazed over. But I'll try to make it interesting.

There are many answers to the question "How many decks?", depending on the faction and whether you are using reinforcements and/or Mercenaries. I'll make this initial discussion manageable, and limit it to Champions, illustrating it mainly with one of the earliest factions, the Tundra Orcs.

Away we go!

**Some Basics**

First, let's review the rules for a Summoner Wars deck:

- A deck consists of 1 Summoner, 1 Reference Card, 9 Event Cards, 3 Wall Cards, 3 Champion Units and 18 Common Units.
- The choice of Summoner specifies the Events, Reference Card and the Units that start on the Battlefield. The remaining Units may be chosen, according to the following rules.
- Each Champion and Common must belong to either the same faction as the Summoner, or to the Mercenaries.
- There may be no more than 1 copy of each Champion, and no more than 10 copies of any kind of Common Unit.
- There may be no more than 6 Mercenary Units total, Champions plus Commons; except if the Summoner is a Mercenary, then all his Units must be Mercenaries.

Now, here are some basic terms of Combinatorial Theory, the branch of mathematics that counts how many ways things can happen.

A **permutation** is an ordering of objects. For example, assuming we summon all 3 Champions in a game, how many orders can they appear in? There are 3 possibilities for the first one summoned, then 2 possibilities for the second one, and only one left for the third:

The number of permutations of N objects is

**P( N ) = N!**

N! is pronounced "N factorial", which is the product of every integer from N down to 1. If N=3, then 3! = 3*2*1=6, as we see above.

## More Counting of Champions

But this is all the same set of Champions. What we want to know is the number of **combinations**, that is the number of distinct sets of Champions. If all we have is the starter set, then we can only do 1 combination because we only have 3 choices to fill in the 3 slots. But if we get Rukor's Power, then we have 3 more Champions to choose from.

Following the same pattern, we have 6 choices for the 1st champion, 5 for the 2nd, and 4 for the 3rd, so 6*5*4=120. (No, I am not going to show all 120 of them.) This gives us the number of permutations of 3 objects out of 6 possible choices.

In general, the number of ways of ordering K objects out of a set of N distinct objects is

**P( N, K ) = N! / (N-K)!**

For the Tundra Orcs -- and most other factions, that is 6!/3! = 6*5*4*3*2*1/(3*2*1) = 6*5*4 = 120. For the Jungle Elves who have 4 champions, we get 4*3*2=24; for the Mercenaries, who have 18 published Champions, we get 18*17*16 =4896.

But again, we want combinations, not permutations, so we divide by the number of ways of permuting the K objects around. The number of K-combinations chosen out of N possibilities is:

**C( N, K ) = N! / (N-K)!K!**

So, since most factions have 6 Champions, we have 6*5*4 / 3*2*1 = 20 ways of selecting 3 Champions from their faction:

The Jungle Elves and Fallen Kingdom have only 4 Champions each, yielding 4*3*2 / 3*2*1 = 4 ways of picking 3 of them:

The Mercenaries, on the other hand, have 18 Champions, so we have 18*17*16 / 3*2*1= 816 combinations:

The Filth have 3 Champions, giving us 3*2*1 / 3*2*1 = 1 combination, which makes sense: there is only one way to choose 3 items out of 3:

Counting only their starter deck, the Filth have only 1 Champion, and there's no way of choosing 3 different Champions out of a group of 1:

So it's good to point out that

**if N<K, C( N, K ) = 0.**

Of course, the Filth don't need as many Champions since they have so many Mutations in their deck. We won't actually count their decks until a later article.

This is enough for one post. We now know how many ways there are for choosing 3 Champions for each faction, as long as we don't mix in Mercenaries. In the next installment we'll see how many ways there are of choosing 18 Commons, which is more complex for several reasons.

# Comments

## Barliman commented:

Actually, it's 816! Before these last two Reinforcement Packs, there were only 220 possible combinations -- adding 50% more Champions almost quadrupled the combos. This shows how explosively combinations can grow, as we'll see even more in future installments.*Posted on 2012-10-01.*

## MacBryce commented:

Thanks for the math refresh. Always useful. :-)So mixing in mercs, that would mean that there are (12 * C((6+18),3)) + C(18,3) + (2 * C((4+18),3)) + C((3+18),3) = 29514 champion combinations throughout the game? Nice!

*Posted on 2012-10-02.*

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## Moon_Knight commented:

Looking forward to playing all the merc champs in 813 diferent combos will come back when its done.Posted on 2012-10-01.