- #1

willr12

- 17

- 2

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?