Safe Combination Possibilities

[willr12]

I am a 15 year old who is obsessed with math...anyway, I developed an equation dealing with safe combination possibilities. For example, if you have a safe with 5 digits, and you know there is (for example) a 4 and an 8 in the combination. You aren't sure which position they are in, and they can be rearranged in all different ways. How many possible combinations are there?
The equation i came up with was as follows:
When y=total digits in the sequence and x=number of digits known,
10^(y-x)y!/(y-x)!
will give you the answer. When you know all the digits, that's the same as saying "how many different ways can n items be arranged?" We all know that can be represented by n!, and this equation satisfies that. When you know all the numbers, 10^(y-x) will cancel to 10^0, since x and y are equal. And on the bottom, (y-x)! will cancel to 0!, which again is 1. Therefore you are left with y!.
Also, if you know 0 of the numbers, the equation is simply 10^y, and this equation satisfies that as well. When the amount of digits you know (x) is zero, the 10^(y-x) cancels to 10^y and the bottom ((y-x)!) cancels to y!. There is also a y! on the top, so those cancel out and you are simply left with 10^y.
So, let's say a friend hands you their phone. While trying to deduce their password, they tell you that their 4-number combination contains a 3, a 6 and a 7. How many combinations are there?
10^(y-x)y!/(y-x)!
10^(4-3)4!/(4-3)!
10^1(24)/1!
240/1
=240 possible combinations
I'm not exactly a professional mathematician (at all) but I just wanted some feedback. Are there any flaws?

 

[WWGD]

I am a 15 year old who is obsessed with math...anyway, I developed an equation dealing with safe combination possibilities. For example, if you have a safe with 5 digits, and you know there is (for example) a 4 and an 8 in the combination. You aren't sure which position they are in, and they can be rearranged in all different ways. How many possible combinations are there?
The equation i came up with was as follows:
When y=total digits in the sequence and x=number of digits known,
10^(y-x)y!/(y-x)!
will give you the answer. When you know all the digits, that's the same as saying "how many different ways can n items be arranged?" We all know that can be represented by n!, and this equation satisfies that. When you know all the numbers, 10^(y-x) will cancel to 10^0, since x and y are equal. And on the bottom, (y-x)! will cancel to 0!, which again is 1. Therefore you are left with y!.
Also, if you know 0 of the numbers, the equation is simply 10^y, and this equation satisfies that as well. When the amount of digits you know (x) is zero, the 10^(y-x) cancels to 10^y and the bottom ((y-x)!) cancels to y!. There is also a y! on the top, so those cancel out and you are simply left with 10^y.
So, let's say a friend hands you their phone. While trying to deduce their password, they tell you that their 4-number combination contains a 3, a 6 and a 7. How many combinations are there?
10^(y-x)y!/(y-x)!
10^(4-3)4!/(4-3)!
10^1(24)/1!
240/1
=240 possible combinations
I'm not exactly a professional mathematician (at all) but I just wanted some feedback. Are there any flaws?

You're 15 and still not a professional mathematician ? Get of the couch and go do something useful with your life : ) !

For the 4-number code, another way, just notice that the 4th number can be any of 0,1,2,...,9 , and can be put in either in 1st, 2nd,etc. and then the other 3 numbers can be permuted, also giving 240 as result, if I understood the rules correctly.

For the first, there are 5C2 ways of choosing the spots for the 4,8 , and for each of these,then the numbers in the other spots can be permutted in any way.

 

[WWGD]

Note that my first comment is just a joke.

 

[willr12]

Note that my first comment is just a joke.

Hahah I know:)

 

[CellCoree]

First of all, I want to commend you for your interest and passion for mathematics at such a young age. Your equation and reasoning are both correct, and it is impressive that you were able to come up with it on your own. However, there are a few things to consider in this scenario.

One potential flaw is that the equation assumes that each digit in the combination can only be used once. In reality, some safes allow for repeated digits in the combination, which would change the number of possible combinations.

Another factor to consider is the order of the digits. In your example, the 4, 6, and 7 can be rearranged in any order. However, in some safes, the order of the digits matters and would result in a different number of combinations.

In addition, your equation does not take into account any restrictions on the digits. For example, some safes may not allow certain digits (such as 0 or 1) to be used in the combination, which would also affect the number of possible combinations.

Overall, your equation is a good starting point and can provide a general estimate of the number of combinations. However, it may not accurately reflect the exact number of possibilities in all scenarios. Keep exploring and learning, and perhaps you can come up with an even more precise equation in the future.