Tool to get all combinations of 4 numbers?

[Netsa]

I'm sure there's a website somewhere. I found a few websites, but most of them have a terrible UI. Does anyone know of one, or have a favorite?

It could be useful if they keep this keypad system, and I don't like relying on my memory for this kind of thing.

[*~> BugDay 3 <~*]

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[CloversDrop]

Numbers 0-9, there are 10,000 possible combinations of just 4 of them, good luck on trying them all.

[Pim]

search an AI site like open ai or google bard (both free), works like a charm

[Netsa]

Originally posted by Cloverdrop:

Numbers 0-9, there are 10,000 possible combinations of just 4 of them, good luck on trying them all.

Only 4 numbers are highlighted and none of them can be used more than once (if there are 3 numbers, 1 of them is used twice). So the calculation isn't 10x10x10x10, it's 4x3x2x1.

[psychomantis100]

https://steamcommunity.com/sharedfiles/filedetails/?id=3038316236

[CloversDrop]

Originally posted by Netsa:

Originally posted by Cloverdrop:

Numbers 0-9, there are 10,000 possible combinations of just 4 of them, good luck on trying them all.

Only 4 numbers are highlighted and none of them can be used more than once (if there are 3 numbers, 1 of them is used twice). So the calculation isn't 10x10x10x10, it's 4x3x2x1.

Forgot you can only use them once.

Formula
C(10, 4) = 10! / (4!(10 - 4)!)
C(10, 4) = 10! / (4! * 6!)

Factorials
10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 3,628,800
4! = 4 × 3 × 2 × 1 = 24
6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

Values
C(10, 4) = 3,628,800 / (24 * 720)
C(10, 4) = 3,628,800 / 17,280
C(10, 4) = 210

So, there are 210 different combinations when choosing 4 numbers from the set of numbers 0 to 9 without repetition.

Will still take a while.

[Nk]

https://nquenault.github.io/pd3codehelper/

[Netsa]

Originally posted by Cloverdrop:

*cut for brevity*

You're not choosing 4 numbers from the set of 0 to 9, the numbers you want are already highlighted on the keypad. It's just 24. That's why there's a long mandatory wait time between each attempt on the keypad. Besides an element of realism, it's to stop players from just trying all possible combinations in less than a minute. But even then, the average number of attempts is...

...well, it's not 24. I'm not exactly Mr. Math, but the point is that it won't take you nearly as long as you think depending on your luck.

That's why I wish they didn't go so hard on the ridiculous side quests needed to sidestep these keypads, because they just encourage standing there and bruteforcing, which is boring even with only 24 combinations, but is so much safer on NRFTW that the only reason to do otherwise is for challenges.

[sk00fy]

This is far and away the best one I've seen:
https://savagecore.github.io/pd3-vault-cracker/

[Netsa]

Originally posted by Xterrestrea1:

This is far and away the best one I've seen:
https://savagecore.github.io/pd3-vault-cracker/

Looks like both of the first two guys suggested that one, but got c-blocked by the automated checking system. I haven't tried it yet, but if it's getting this many recommendations, I'll just assume it's good.

[Prussian Wolf]

Originally posted by Cloverdrop:

Originally posted by Netsa:

Only 4 numbers are highlighted and none of them can be used more than once (if there are 3 numbers, 1 of them is used twice). So the calculation isn't 10x10x10x10, it's 4x3x2x1.

Forgot you can only use them once.

Formula
C(10, 4) = 10! / (4!(10 - 4)!)
C(10, 4) = 10! / (4! * 6!)

Factorials
10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 3,628,800
4! = 4 × 3 × 2 × 1 = 24
6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

Values
C(10, 4) = 3,628,800 / (24 * 720)
C(10, 4) = 3,628,800 / 17,280
C(10, 4) = 210

So, there are 210 different combinations when choosing 4 numbers from the set of numbers 0 to 9 without repetition.

Will still take a while.

You don't randomly put them in though, you get highlighted numbers from the keypad.
You get 2-4 numbers and they can repeat, thus you can have from 14 to 36 codes.

[Unknown Knight]

In before new security modifier: All employees wear gloves to prevent oil stains and markings on keypads.

[Unknown Knight]

Looking at a permutation formula, it seems to be 6/12/24 permutations for 2/3/4 unique numbers.

Edit: Oh I see where I went wrong. 14/36/24 is right.

[Netsa]

Okay, here's my count of 2 and 3 number combos:
1222 2111
1211 2122
1212 2121
1221 2112
1121 2212
1122 2211
1112 2221
14 for 2-number combinations.

1123
1132
1232
1223
1233
1213
1231
1312
1323
1332
1322
1321
Multiply by 3, 36 for 3-number combinations.

No idea what the calculation is on these, but writing them out by hand got weird.

[Nk]

Number of combinations :
14 for 2 fingerprints
36 for 3 fingerprints
24 for 4 fingerprints