How many possible spell combinations for a wizard brawling game?

Question

I'm working on a game that involves drafting different spells and I'm trying to figure out how many combinations are possible.

Simply put, each spell has an element and a class:

• There are $10$ elements (Fire, Water, Air, Earth, etc.)
• There are $7$ different classes (Primary, Movement, Melee, Defensive, etc.)
• There is one spell for each element/class combination, therefore $70$ spells
• By the end of the game a wizard will have $7$ spells: one of each class

My questions are:

1. I believe there are $10^7$ combinations if there are no element restrictions. Is this correct?

2. If the drafting was restricted to a random $4$ of the $10$ elements each game, how many possible combinations would there be?

   EDIT: You would still be drafting $7$ classes of spells, but any combinations with more than $4$ different elements would be illegal. A single game would have $4^7$ combinations, but I'm wondering about all possible games using $4$ out of $10$ elements.

Answer 1

I believe there are $10^7$ combinations if there are no element restrictions. Is this correct?

Yes.

If the drafting was restricted to a random 4 of the 10 elements each game, how many possible combinations would there be?

First, choose which four elements are drafted, then choose a spell of each class in the allowable elements. That's $\binom{10}{4}\cdot 4^7$.

But wait - we've overcounted. If we only used three of the elements in our final spell list, then there's no way of telling which other element was drafted but not used. There are seven possibilities, so those are counted seven times each, and we'll need to compensate. Subtract six times the number of ways as if we drafted three elements: $-6\cdot\binom{10}{3}\cdot 3^7$.

Then, this repeats for two-element spell lists. We've counted them $\binom{8}{2}=28$ ways from four-element drafts and $-6\cdot 8=-48$ ways from three-element drafts. To compensate, we add back $21$ times the number of ways as if we drafted two elements: $21\cdot\binom{10}{2}\cdot 2^7$.

Finally, one-element spell lists. They're counted $\binom{9}{3} = 84$ ways from four element drafts, $-6\cdot\binom{9}{2}=-216$ ways from three-element drafts, and $21\cdot 9=189$ ways from two-element drafts. To compensate, we subtract $189-216+84-1=-56$ times the number of wats as if we drafted one element: $-56\cdot\binom{10}{1}\cdot 1^7$.

Overall, that's $$\binom{10}{4}\cdot 4^7 -6\cdot\binom{10}{3}\cdot 3^7 + 21\cdot\binom{10}{2}\cdot 2^7 -56\cdot\binom{10}{1}\cdot 1^7 = 1986400$$ valid spell lists. Plenty of variety to work with.