The number of 5-digit numbers such that their sum of digits is even is? I tried this problem by counting the number of 5-digit numbers and then dividing it by 2 just like we do for three digits. Please help. Thank you.
 A: In how many ways can the five digit number be chosen so that the sum of digits is even? The first digit can be chosen in 9 ways (since it can't be a 0), the second in 10 ways, the third in 10 ways, the fourth in 10 ways, and the fifth in five ways (the last digit must be one of the five even digits if the first four digits add up to an even number, and odd otherwise, so that the total sum is even).  This gives $9 \times 1000 \times 5 = 45000$, which is half the number of 5 digit numbers. 
A: Can you see how many $5$ digit nos. can be formed with nos. $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$? 
Clearly $1$st place can't be occupied by $0$, so the $1$st place can be filled from $9$ different numbers.   The rest of the places can be filled from $10$ different numbers, so the total number of five digit numbers that can be formed is $9\times10\times10\times10\times10$. 
Now these many digits will be either even or odd after summation.  But, we are interested in the numbers whose sum is even so it will be half of the total five digit numbers, so the answer is $\frac{1}{2}\times(9\times10\times10\times10\times10)$ which is $45000$.
A: There are $90000$ $5$-digit numbers, exactly half have an even sum of digits (base $10$ has equal numbers of even and odd digits) 
A: Let $S$ be the set in question. Let $E$ be its complement in the set of all  5 digit numbers. $S $ and $E$ partition the collection of all 5 digit numbers and $S$ does not contain 99999. Adding 1 to elements of S gives elements of E. This function is from S to E and is in fact a bijection and so both sets contain equal number of elements. So you only need to count how many 5 digit numbers are there.
