So, in a game I play, we do alliance missions, and we are in alliances of $18$ people. I was thinking of writing a computer program to go through all the combinations to figure out the optimal placement of all our members.
We get divided into three battlegroups of $6$ each, and the order doesn't matter, so from a little research on the internet, I found out that that would be $\frac{18!}{6! \times 6! \times 6! \times 3!}=2858856$ combinations of just putting $18$ members into $3 $groups.
Once in a battle group, and you begin the mission, there's immediately two ways to go, right or left. So that gives $\frac{6!}{3! \times 3!}=20$ combinations of $6$ people, $3$ going right and $3$ going left.
That gives us $57177120$ combinations at this point.
Once going right or left, each of the groups of three, let's call them players A, B, and C, can pick one of three paths that then converge at a mini-boss. The same thing three times, three sets of three paths, three mini-bosses each of which generally one player takes. Am I right to think this makes $6\times6\times6\times6=1296$ combinations for each side of each battlegroup?
An example of my side of my battlegroup is as follows:
Player A takes paths $1, 1,$ and $3$ and mini-boss $3$
Player B takes paths $3, 3,$ and $1$ and mini-boss $1$
Player C takes paths $2, 2,$ and $2$ and mini-boss $2$
So then am I correct in that there are a total of $74101547520$ combinations for the computer program to test?
You multiply by $20$ only one time ... but since there are three battle groups, you need to multiply by $20^3$
Same for the paths at the end: you would need to multiply by $1296^6$
Too big for a computer to handle ... so I suggest focusing on some particular aspect, e.g. Just the ways in which to make the $6$ subgroups