Given:
A 4-digit number contains at least one even digit.
To find:
How many such numbers are there?
Solution:
Digits are 0,1,2,3,4,5,6,7,8,9.
Number of odd digits = 5
Number of even digits = 5
4-digit numbers are from 1000 to 9999.
Total number of 4-digit numbers = 9000
Possible ways to get odd digits on all the 4 places is
[tex]5\times 5\times 5\times 5=625[/tex]
It means 625 numbers are their in which all 4 digits are odd.
To find total 4-digit number that contains at least one even digit, subtract the numbers which contains only odd numbers from the total 4-digit numbers.
[tex]9000-625=8375[/tex]
Therefore, total 4-digit number that contains at least one even digit is 8375.
