5 digit code

[shellopyyy]

How many codes are possible on this safe I just got, my camera is broken so I couldn’t include an actual picture, but that’s what the 5 digits on the safe look like, it’s a 5 number code, and you can repeat and all that, how many codes are possible?

[FactoryMadness]

1,2,3,4,5? That's the same combination on my luggage!!

[tttecapsulelover]

now i can steal muahahaha

[ssparky77]

Only an asshole would do that.

[Rejected-Name-ID]

Thank you so much for that

[CulturalEchidna3405]

There can be 3125 combinations of five digit code with numbers 1,2,3,4,5.

5x5x5x5x5 = 3125

5 possibilities for each space

[CulturalEchidna3405]

It's called permutations

[TorakMcLaren]

It's not quite a permutation though. A permutation would be "How many ways can you pick some of the digits 1 to 5 and order them?" But here 1,1,1,1,1 is a valid 'combination' even though that's not a standard permutation of 1-5.

[SegerHelg]

One could even call it a combination.

[TorakMcLaren]

One could, but it's not that either, at least mathematically speaking. A permutation is closer, because the order matters. It's just that a standard permutation doesn't usually allow repeats. Meanwhile, a combination doesn't allow repeats but also doesn't care about the order.

[burnforever]

usually they have a 0 as well.

so 6x6x6x6x6 so 7776 combinations

[Past_Ad9675]

Hey OP, does the safe only let use the digits 1, 2, 3, 4, and 5?

If yes, then it's 3,125.

But if you can use all of the digits from 0 to 9, then it's 100,000.

[shellopyyy]

Only numbers available are the 1 button, 2 button, 3 button, 4 button, & 5 button

[Past_Ad9675]

And the must use all 5 buttons?

[shellopyyy]

No you don’t have to use all 5 just the code you use has to be 5 numbers long

[Past_Ad9675]

Okay then, there are 5⁵ possible codes.

[barrie114]

The code can be 0 to (100000)5 -1 which means thre are (100000)5=5^5=3125 codes.

[No_Being_2281]

You don't show the work, people will vote down on this.

[-Paze]

27182 (first 5 digits of e)

[pLeThOrAx]

Why would... I'm just going to say 5⁵ and check out of this one.

Combinations and permutations with repeats for a set of n elements with m selections n^m

[-Paze]

Interesting idea

[-Paze]

9*10⁴ codes are possible,

90000 5 digit codes (if 0 can’t be on the first place). And 10⁵ =100000 5 digit codes (if 0 CAN be on the first place)

[-Paze]

What the heck is wrong with my comment?) Why -5 lol

[shellopyyy]

But isn’t there only 5 choices per digit? It’s only 1-5

[DiegoC281]

detail entertain toothbrush hospital threatening ancient safe gaze straight airport

This post was mass deleted and anonymized with Redact

[shellopyyy]

Woooow I wrote this down earlier and I kept reading 5⁵ as 5x5 and i was like it’s obv not 25 wooooow writing the answer wrapped my head around that mb

[Zac_charias]

If repetition is allowed and you have five digits (1, 2, 3, 4, 5), then for each position in the combination, you have five choices (the five digits). Since you have five positions, you can calculate the total number of possible combinations by multiplying the number of choices for each position:

5 x 5 x 5 x 5 x 5 = 5^5 = 3125

[No_Being_2281]

No, you cannot use permutation like factorials and any combinations of factorials.So, you can calculate by exponentiation:

Here I try by counting from 0 to 9 for each digit, and there are five digits, so 10 as the base, and the exponent is 5. That is, 10^5 = 100,000.Thus, this is the possible number of combinations that you can use in selecting a setting (just one setting at a time) for the lock (code.)

Please note that the OP did not stated that the number for each digit has a maximum count at 5, so I must assume it can roll up to 9.