How many 4 digit integers using digits 1-9 are there so that the sum of digits is even?

Question

How many 4 digit integers using digits 1-9 are there so that the sum of digits is even?

I divide it into cases.

• Case 1, all even: $4^4$
• Case 2, no even: $5^4$
• Case 3, two even, two odd. How should this be done?

Answer 1

For case 3:

• Pick which two of the four digits are even: $\binom42=6$
• Assign the digits: $5^2\cdot4^2$

There are $6\cdot5^2\cdot4^2$ integers in this case. The total number of integers is $4^4+5^4+6\cdot5^2\cdot4^2=3281$.

Answer 2

You need to choose two (out of four) of the digits to be odd, which you can do in ${}^4 C_2$ ways. Once you've decided which digits are going to be odd and which are going to be even, you can just select the digits in order, for which you have $4^2 5^2$ choices.