Answer:
5040,56
Step-by-step explanation:
We have to construct pass words of 4 digits
a) None of the digits can be repeated
We have total digits as 0 to 9.
4 digits can be selected form these 10 in 10P4 ways (since order matters in numbers)
No of passwords = 10P4
=
b) start with 5 and end in even digit
Here we restrain the choices by putting conditions
I digit is compulsorily 5 and hence only one way
Last digit can be any one of 0,2,4,6,8 hence 5 ways
Once first and last selected remaining 2 digits can be selected from remaining 8 digits in 8P2 ways (order counts here)
=56
