Answer

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**Hint:**

Firstly, we need to find the total number of 7-digit numbers. This we can find by calculating the number of permutations at each digit’s place. The numbers with sum of digits as even will be the half of total 7-digit numbers.

**Complete step by step solution:**

We know that there are 7 digit places in a 7-digit number. In each place except the ${1}^{\text{st}}$ digit place, numbers from 0 to 9 can be used. For a number to be a 7-digit number at the ${1}^{\text{st}}$ digit place numbers from 1 to 9 can be used (because using 0 at ${1}^{\text{st}}$ digit place will make the number a 6-digit or lesser number). So, we can obtain the number of all possible seven digit numbers by multiplying all possible numbers at each digit’s place.

Thus, the all possible 7-digit numbers is given by

$

9 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10 \\

\Rightarrow 9 \times {10^6} \\

$

Now, half of these numbers have the sum of their digits as an odd number. So, the other half of these numbers have their sum of digits as an even number.

$

\dfrac{1}{2} \times 9 \times {10^6} \\

\Rightarrow 45 \times {10^5} \\

$

**Therefore, the correct answer is B.**

**Note:**

The concept of half of numbers as an even number or an odd number is valid for any number of digits. We can use permutations to calculate the total numbers for any number of digits. Permutation is always used to get the arrangement of items.

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