- #1
fomenkoa
- 47
- 0
Hey everyone
I have this for Discrete homework:
A lock has the numbers from 0 to 59 ...A combo is made up of 3 numbers...How many combos are possible if the 2nd and 3rd numbers have to differ by at least 3
The answer is whatever 60 times 58 times 57 is
I know why the 60 is there..because the 1st number can be any of the 60. How they got 58 times 57 I'm not exactly sure...
Can anyone help?
Anton
I have this for Discrete homework:
A lock has the numbers from 0 to 59 ...A combo is made up of 3 numbers...How many combos are possible if the 2nd and 3rd numbers have to differ by at least 3
The answer is whatever 60 times 58 times 57 is
I know why the 60 is there..because the 1st number can be any of the 60. How they got 58 times 57 I'm not exactly sure...
Can anyone help?
Anton