How many 4-digit numbers have the second digit even and the fourth digit at least twice the second digit?
Please expalin/give me a hint on how to do it! Thanks!
Currently, I have 9*9*2*8 as there are 9 ways for the 1st and 3rd digit, 2 ways for the second digit, and 8 ways for the last one. This is incorrect and I do not know why so it would be nice if you could explain why im incorrect