Simple (possibly?) Problem

>>8480729
5! = 5*4*3*2*1 = 120
/thread

>>8480731

Could you explain the reasoning being this equation please (as simply as you can)?

>>8480740
Not anon who replied. But you can pick from five potions to begin with. There are five ways to do this, one for each potion. Then you have four potions left to pick from. There are four ways to do this for each previous potion. So there are now 5*4 ways to pick a potion.

You now continue till you run out of potions, which makes for 5! Ways to combine the potions.

>>8480752

Awesome. Thank You! I imagine this is a slow board so I'm going to leave this thread open in case anyone has fun answers for my last question (about related puzzles), but I'll try to delete it if that'd be preferable.

>>8480761
Two complications apply to >>8480752 .

First, 5*4*3*2*1 is the number of possible orders in which you can select the 5 liquids. But this is not quite the number of possible mixtures. If you choose order [red, green, purple, blue, yellow], start by mixing red and green, and then add in purple, blue, and yellow in order -- does that give you a different potion than if you choose [green, red, purple, blue, yellow]? In that case, you ALSO start by mixing green and red. Is that a different potion? (If the first ingredient has to simmer for five minutes in a cauldron before adding the second, the answer is probably "yes". If it's just a matter of mixing, the answer is probably "no".) If not, divide >>8480752's number by two.

Second, can you for example do the following:
- mix red and green.
- mix yellow into the red/green.
- mix blue and purple.
- mix the red/green/yellow with the blue/purple.
If all the mixing needs to happen in a single cauldron, probably not. If you can just mix vials together, probably so.
If so, that yields a different number than >>8480752 said. This one is a bit harder to compute in detail.

>>8480774
And if you want to be even more completionist you can have 3 or more colors mixed simultaneously. And have 3 or more mixed batches mixed simultaneously.

>>8480774
>This one is a bit harder to compute in detail.
Lazy way: Compute the first few numbers, then look up in OEIS. Here it is:
http://oeis.org/A001147

>>8480774
Ah, yeah, I assumed OP was just having them mix in one potion at a time and that the order they were mixed in mattered, so that there would be only 120 different potions.

>>8480852
Now try this one >>8480841

OP here; it seems like this question is a lot more complicated than intended.

I'm going to try to simplify the problem for the sake of simplifying it for myself.

1) A single person drinks a poison.

2) They have 5 liquids in vials laid out before them.

3) They must mix the 5 liquids in the correct order (a,b,c,d,e) in order to concoct an antidote.

I'm note sure how to word this in "mathematical terms", but I'm looking for a simple 1+2+3+4+5=antidote equation. Would I be better off (under the assumption that my PCs are of average intelligence) going with a 3 liquid equation?

>>8480916
No, the answer is simply 4! = 24 in that case. But trivial problems are no fun.

>>8480922
Typo? Shouldn't that be 5!, or 5!/2 if the order of the first two doesn't matter?

>>8480875
This I'm not finding in the OEIS. Could be not there or I'm miscomputing the terms.

I get for the first few terms
1,1,4,25,201,...

Breakdown of the 201:
(((AB)C)D)E 60
((AB)C)(DE) 30
((AB)(CD))E 15
(ABC)(DE) 10
((AB)CD)E 30
((AB)C)DE 30
(ABC)DE 10
(ABCD)E 5
(AB)CDE 10
ABCDE 1

>>8480922

Thanks, I now (think I) understand the correct equation to answer this type of problem.

I understand that this type of problem might not be fun, but they're sometimes neccesary in a d&d campaign. Do you guys have any "fun" math problems that you could turn into a D&D mission?

>>8480933
25 was a mistake, should be
1,1,4,26,201,...

ABCD
((AB)C)D 12
(AB)(CD) 3
(ABC)D 4
(AB)CD 6
ABCD 1

Still no OEIS results.

>>8480941
http://oeis.org/A000311
looks like the answer, which means my 201 is missing something.

OP here, once again (sorry).

Would a (3) liquid equation be 3*2*1 resulting in (18) different results?

Also, just to make sure this thread isn't a complete waste, do you guys have any other equations that you would consider "fun enough to incorporate into a D&D quest"?

>>8480948
screw all the complications, just do it based on order alone, which for 5 liquids gives 120 combos, or 4 liquids gives 24 combos

throw a monty hall problem or three utility problem for luls? ask them to find non trivial solutions to the riemann zeta function?

ask them to do a finite quantum walk with multiple absorbing boundaries? I dunno

>>8480943
was missing
((ABC)D)E 20
(AB)(CD)E 15

>>8480948
3*2*1 = 6
Do you need your players to compute the number of possibilities or try them all?
In either case, you'd need to be very clear about what the rules are for how you mix stuff. If it works out to just counting the number of orders, that's something normally taught in high school, although many people will have forgotten.

>>8480729
You're going to have at least one player and probably 3 or 4 who will immediately know this. Might want to go for something harder.

>>8480959

The "solution" of the problem would simpy be mixing the liquids in the right order; either through an equation or random mixing.

They will lose a percentage of HP for every "wrong" mixture.