How many 4-digit numbers are there where any two consecutive digits have different parity?

Guest May 7, 2022

#1**0 **

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.

Guest May 7, 2022