# Moderate Permutation-Combination Solved QuestionAptitude Discussion

 Q. How many 5-digit positive integers exist the sum of whose digits are odd?
 ✖ A. $36,000$ ✖ B. $38,000$ ✔ C. $45,000$ ✖ D. $90,000$

Solution:
Option(C) is correct

There are $9 × 10^4 = 90,000$, 5-digit positive integers.

Out of these $90,000$ positive integers, the sum of the digits of half of the numbers will add up to an odd number and the remaining half will add up to an even number.

Hence, there are $\dfrac{90,000}{2} = \textbf{45,000}$ 5-digit positive integers whose sum add up to an odd number.

Edit: For an alternative approach, check comment by Carter.

## (4) Comment(s)

Snehal
()

A common sense approach can be (max 5 digit num + min 5 digit num+1) devided by 2

=(99999-100000+1)/2

Snehal
()

correction : (max 5 digit num - min 5 digit num +1)/2

=> (99999 - 10000 +1)/2 = 45000

Vince
()

positive integers should be 10*10*10*10*5 since the last digit should be even(0,2,4,6,8) =50000 so 25000 should be the answer.

Carter
()

5 digit positive number can not begin with 0, so 9 choices for first digit and 10 choices (0 to 9) for the remaining digits, so $9*10*10*10*10 = 90000$.