How many passwords of 4 digits are there(a)if none of the digits can be repeated? (b)start with 5 and end in an even digit?

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

120 seems like the most sensible answer.

step-by-step explanation:

if there is only one topping on all caramel apples and there's 120 of them. then all apples should/will be covered in sprinkles. make sense?