Question

The number of 7-digit numbers the sum of whose digits is even is
\[
  A.35{\text{ }}x{\text{ }}{10^5} \\
  B.45{\text{ }}x{\text{ }}{10^5} \\
  C.50{\text{ }}x{\text{ }}{10^5} \\
  D.None\,\,of\,these \\
\]

Answer 1

Hint: We need to find the total number of 7 digit numbers possible. This can be found by permutations of the numbers that can have in each digits place. The numbers with the sum of digits is even half of the total number of 7digit numbers.

Complete step-by-step answer:
In a 7 digit number there are seven places. In each of the seven places except the 1st digit, numbers from 0 to 9 can be used. But in the 1st digit’s place , the numbers except 0 can be used. That means we can use numbers 1 to 9. This is because, if we have zero in that place, the number will not be a 7 digit number. It will become a number of 6 or lesser digits. So, the number of possible seven digits numbers can be calculated by multiplying the number of possible numbers at each digit’s place.
Thus, we get, total number of 7 digit number = \[9{\text{ }}X{\text{ }}10{\text{ }}X{\text{ }}10{\text{ }}X{\text{ }}10{\text{ }}X{\text{ }}10{\text{ }}X{\text{ }}10{\text{ }}X{\text{ }}10{\text{ }} = {\text{ }}9{\text{ }}X{\text{ }}{10^6}\]
The number of 7-digit numbers having sum of digits as an odd number is the half of the total 7 digit number.
So, the number of 7-digit numbers whose the sum of digits is even is $ = \dfrac{1}{2} \times 9 \times {10^6} = 45 \times {10^5}$
Therefore, the correct answer is option B

Note: The concept that half of the total number have the sum of digits as odd and other half have sum of digits as even is valid for every particular digit number. The concept of permutations is used to find the total number of 7-digit numbers.