Question

Consider a locker that requires a code to open. For simplicity, assume it only uses digits 0 through 9. To unlock it, you need to use a 3-digit code that uses distinct digits, but you've forgotten your code.

a) How many different codes (maximum) will you have to try to open the locker?

b) Consider the same locker in part a). How many different codes will you have to try if you remember the second number is 9?

Answer 1

It says 'distinct digits' and 3 digit code.

a) Total different codes = 10*9*8 = 720

As we need to use distinct digits so the one used in first place cannot be used further that's why ways are reduced by 1 in each step.

b) If the second number is 9 then total different codes = 9 * 1 * 8 = 72

As '9' is already used so we cannot use it again at anyother place. That's why counting started from 9 for first place then 9 which is fixed so 1 way and then 8 ways for last place.

I hope this will help. Thank you.