What is the Maximum number of ATM PINs that ever exist?

Befor we understand how we can know all the possible number of ATM PINs or 4 digit passwords, we need to understand a little bit of probability in the area of permutations and combinations.
Permutations are used in statistics to determine the number of ways a set of things can be arranged while combinations are used to determine the number of ways you can choose from a set of things .
For instance you can have a set of 10 students, you can use permutation to determine how many ways you can arrange these 10 students taking 5 students at a time ,6 students at a time, etc , but i'll explain that later in this article .Combinations on the other hand are used to determine the number of ways you can choose from a given sample at a time.

So at this point you might be wondering the difference between "Choose" and "Arrange".
Lets assume you have a given set of things A,B,C,D. If you want to choose from the set of given things taking two items at a time you can have A,B or C,B and when you choose A ,B = B,A but when you arrange A,B is diferent from B,A which means A,B≠B,A (Read as A,B is not equal to B,A incase you are not familiar with mathematical terms) .In this article im going to focus more on permutations.

 

What are Permutations?

As i said earlier permutations are used in statistics(probability) to arrange things it is represented this way:

 

ⁿP_r

 

Which is read as n Permutation r. If we have 3 items "1,2,3" lets say and we want to calculate the number of ways we can arrange 3 things taking the 3 at a time we can say 3 permutation 3 or simply 3!(read as 3 factorial ) .

n!(read as n factorial) can be writen as:

 

n*(n-1)(n-2)(n-3)*......*1

Well if you are not familiar with math the above formula might seem a little bit complicated but you shouldn't worry about that , i'll explain it a little bit befor we move forward. If you have three things that are different, "1,2,3 " lets say you can arrange them in 3! (three factorial) ways ,that is

3! → 3*2*1 = 6
4! → 4*3*2*1 = 24
5! → 5*4*3*2*1 =120
....... and so on

So in the case of 3! or number of ways we can arrange three things we can say 3! → 3*2*1 = 6 so there are 6 ways of arranging 3 things that are different. so in the case of "1,2,3" we can arrange them this way :

1,2,3
3,2,1
1,3,2
2,1,3
2,3,1
3,1,2

so we where correct! the maximum number of ways you can arrange three things are 6 ways or 3! ways.
Note this condition does not hold if 2 things in the sample we are given to arrange repeat, for instance 1,3,3 can be arranged in 3!÷2! not 3! ways because 2 things repeat , that is 3 and 3.

Okay now we move to permutations,if we are given 10 things to arrange taking 4 at a time , we cannot say 10! because 10! will give us the number of ways we can arrange the 10 things taking the 10 at a time ,this is where permutation comes in.so in a case where we are looking to arrange 10 things taking 4 at a time we will say 10 permutation 4 which is written as :

 

¹⁰P₄

 

Note that the higher number always comes firs when we are reprrsenting permutation in this form ,that is ,the value of n is always greater than or equal to r .

 

Lets explain this better if we have ⁿP_r

 

We can calculate it using the formula

 

 n!

---

 (n-r)!

 

So in our case we wanted to calculate ¹⁰P₄ (read as 10 permutation 4) we will subtitute the values of n and r into the equaion to have

 

10!

---

 (10-4)!

 

which gives us 5040, This means that there are 5040 of arranging 10 things taking 4 at a time.

 

How to derive the all the ATM pins

There are 10 digits that 4 digit pins can be derived from :
0,1,2,3,4,5,6,7,8,9
Lets start from the simplest .We can have 4 digit passwords like:
0000,1111,2222,3333,4444,
5555,6666,7777,8888,9999
Which means there are 10 possible passwords of this type. Next we can have passwords like
1122,2233,4455
and so on just by taking 2 numbers at a time and making them 4 digits.In this case we are taking 2 things out of 10 things at a time and making them 4 by rewriting each of them one more time.For instance we can take 1 and 2 out of 0-9 and make it 4 digits like this :
1122 or 2211 or 1212 or 2121
So how do we find the number of 4 digit passwords of this kind? From 0-9 (10 things) we can take 2 digits at a time .The number of ways we can arrange 10 things taking 2 at a time is ¹⁰P₂ which gives 90. But we also need to make them 4 by writing 1122,2211, etc. Next we need to arrange 1122,2211 so how to we do that we say:

  4!

---

2!2! Which gives 6 ,that means we can arrange 4 thing in 6 ways if 2 things in the sample are identical .So for 4 digit passwords which 2 numbers are identical there are 6*90or 6(¹⁰P₂) possible passwords.

We can also have passwords like
1123,3312,9811 .That means out of 4 digits 2 digits are identical ,the other 2 are different. So how many type of this passwords exist ? first we need to understand that in this case , out of 0-9 we are taking 3 numbers at a time so we .We can take 3 things out of 10 things at a time in ¹⁰P₃ ways. But there are 4 digits so we also need to calculate the number of ways we can arrange 4 things where 2 things are identical.We can do this by :

 

  4!

---

2!

 

So there are 12(¹⁰P₃) possible types of this password.

We can also have passwords that nothing repeats ,that is , the 4 digits are unique and there are no identical digits. Since there are no identical digits ,there are ¹⁰P₄types of such passwords.

Finally we can have passwords like 1113, 2221,5000 ,etc. Just passwords where 3 digits repeat . There are 10P2 maximum number combinations that could makeup such passwords. But you can rearrange those passwords formed in 10P2 ways by 4!/3! Ways . So the maximum combination of such passwords will be  (4!3!)(10P2)=360

Now its time to add up everything to find the total possible number of 4 digits passwords or ATM pins.

10 + 10P4 + 6(10P2) + 12(10P3)+4(10P2)=14230

 

So we have 14,230 possible ATM pins!.

Share your thoughts in the comment of you doubt my calculations