How many three digits even numbers can we form such that if one of digit is $5$ the following digit must be $ 7$?
I need some ideas on how to proceed on this problem.
How many three digits even numbers can we form such that if one of digit is $5$ the following digit must be $ 7$?
I need some ideas on how to proceed on this problem.
You have two different kinds of such three-digit even numbers.
$57x$, where $x$ can only be $0,2,4,6,8$ which is just $5$ possibilities.
For the remaining, you count all even three-digit numbers with no $5$ in them. This will be $8\times 9\times 5 = 360$
So you have $365$ possibilities.
required number of ways = number of ways is number of ways of forming a 3 digit even number(10.10.5)- number of even numbers in which 5 is the first digit and 7 is not the second digit(1.9.5) = 455