LetMeEatCake

Yes, it matters. If you're picking 1 out of 10 each from 10 different sets, you get 100 combinations. This also limits the sample space to what is possible.

For simplicity's sake so we can do math that we can intuitively figure out, look at it as picking one from binary choices, with three companions. So you have companions A, B, and C. With possible endings A1, A2, B1, B2, C1, C2.

If you pick 1 from A, 1 from B, and 1 from C you get 2*2*2 possible outcomes, or 8.
If you pick any 3 from the set of 6 (A1, A2, B1, B2, C1, C2) you get 6!/(3!*3!) possible outcomes, or 20.

With the former, you always get one ending for each companion. Every companion has an option selected, and every companion does not have multiple endings selected. With the latter, you might get 1 from each companion. Or you might get A1, A2, and B1 — with no endings for companion C, and two endings for companion A.

How can ending A1 "A lived happily ever after" and ending A2 "A died midway through the player's journey, never having found happiness" both happen? They cannot. We need to use a system that limits the sample space to exactly 1 per companion, even if that option itself might be "doesn't show up in the end slides."

Unless I'm getting the math wrong myself, for any "pick 1" combination set like this we're dealing with just multiplying the combination sets together. Technically we're multiplying by the factorial of the sample size, but 1!=1.

We're not picking any 10 from within the subset of 100; you cannot pick both ending 1 and ending 4 from companion A and then no ending at all for companion C. I'm assuming each individual sub-ending is mutually exclusive with the rest of its sample space. That difference of assumptions is what led to your 1.7x10^13^ combinations.

I agree and I suspect companions are carrying a lot of the weight for this calculation.

Hypothetically, if there's 10 companions with 10 individual endings each you'd get 100 endings right there. Add in 10 main endings and you get 1000, add in 4 major side quests and 4 variations each and you're at 16,000 ending variations.

Yes. In fact, I'd say that Firefox runs clearly better than Chrome does these days. An inversion of the past.

Awesome, thanks for the help!

I'm hoping to relocate to the Boston area sometime Soon^TM^ and this is a great future resource for me!

With respect to rental agents — is this something where individuals have to seek them out ourselves, or will the landlord of any particular rental unit point applicants towards their preferred agent/agency?

No idea, sorry.

This would be good, and mitigating climate change is going to require a solution by a thousand cuts.
To put the 230 megatons in context, global CO2 emissions in 2022 are estimated at ~40 gigatons, and total GHG emissions at ~50 gigatons. This would represent a ~0.5% reduction in emissions. Not bad at all, all told.

Yes. Thunderbolt 5 supports DP 2.1, with up to 80 Gbit/s of bandwidth. A 5k/120 with 10 bit color and no DSC display needs 57 Gbit/s of bandwidth.

DisplayPort 2.0 can 5k/120 with DSC and UHBR 10 cables, or 5k/120 without DSC on UHBR 13.5 cables. HDMI 2.1 can handle 5k/120 with DSC in use.

It's not a bandwidth limitation.

If they concluded that they could raise prices to increase profit, they'd do so regardless of theft rates. Those are separate issues.

"Used to be" being the operative term. The GOP you speak of is dead, has been dead for many many many years, was dying for many years before it died, and will remain dead until the current party coalitions break down and we see a new alignment of the electorate.