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?
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?
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