If that helps, but I'm pretty sure the answer is -1 that is what I have found in Google that is what all the answers are that in finding.
7/22 you add the denominators toghether and add the numerators.
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
12 total letters...5 vowels and 7 consonants.
probability of 1st letter being a vowel is 5/12
after replacing ...
probability of 2nd letter being a vowel is 5/12
probability of both occurring is (5/12 * 5/12) = 25/144 <===