How many 4-digit numbers are there where any two consecutive digits have different parity?
Here is my attempt:
So suppose the 4 digits are: _ _ _ _
The first _ could be even or odd.
If it is even then,
let Case (1): EVEN ODD EVEN ODD
If it is odd then,
let Case (2): ODD EVEN ODD EVEN
Even numbers: 0,2,4,6,8
Odd numbers: 1,3,5,7,9
For Case (1):
Since the first digit _ could be 2,4,6,8 then there are 4 possibilities.
The second digit is odd, so it could be 1,3,5,7,9 then there are 5 possibilities.
The third digit is even, so it could be 0,2,4,6,8 (Notice, the first digit couldn't be 0, as that would make the number "3-digit") then there are 5 possibilities.
The fourth digit is odd, so it could be 1,3,5,7,9 then there are 5 possibilities.
Thus multiplying all the possibilities:
4*5*5*5 = 500 (*)
For Case (2):
Same method as Case (1), except the first digit is odd so it could be: 1,3,5,7,9 then there are 5 possibilities
5*5*5*5 = 625 (**)
Adding (*) and (**):
Therefore, 500+625 = 1125, 4-digit numbers which satisfy the given condition.