How many 5 digit numbers are there that are even?

Answer

Hint: We have five places, ones, tens, hundredth, thousandth, and ten-thousandth place in any 5 digit number. We will first find the number of 5 digit numbers that can be formed. Then find the number of digits whose sum is even by dividing the total five-digit numbers by 2.Complete step by step answer:We will first calculate the total number of possible 5 digit numbers.We have five places, ones, tens, hundredth, thousandth, and ten-thousandth place in any 5 digit number.Each of these places can be filled with either of the 10 digits except for the ten-thousandth one because if we include 0 in the ten-thousandth place, we will get a four-digit number. So, for one place we have only 9 choices.Also, repetition is allowed for the numbers.Therefore total number of 5 digit numbers can be calculated as $9 \times 10 \times 10 \times 10 \times 10$ which is equal to 90,000.Out of the total digits formed, half of them will have the sum of digits as even and half of them will have sum of digits as odd.Thus, from 90,000 total possible digits, exactly half of it will have sum of digits as even sum.Therefore, divide 90,000 by 2.We have $\dfrac{{90,000}}{2} = 45,000$, 5 digit numbers that have sum of digits as even.Hence, B is the correct option.Note:- While calculating the total number of 5 digits number, do not include 10 in all the places because 0 in ten-thousandth place will give you 4 digit number. Alternatively, we can take 10 digits for each place and then subtract the numbers that are formed when the ten-thousandth place is 0.

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_ _ _ _ 2 Available digits 0,1,3,4,5

Out of these 5 digits leftmost digit can be filed with 1 or 3 or 4 or 5 As if 0 comes on the left most position the number will become a 4 digit number. So, number if ways of filling leftmost digit = 4

As 2 digits are already used, the next position can be filled with either of the 4 digits .

Proceeding in this manner, the next 2 digits can be filled in 3 and 2 ways respectively.

$\underset{4}{\stackrel{-}{|}}\underset{4}{\stackrel{-}{|}}\underset{3}{\stackrel{-}{|}}\underset{2}{\stackrel{\_2}{|}}$

So, the number of possible numbers = $4\times 4\times 3\times 2$ = 96

Case 3: Rightmost digit is 4
This case is similar to 2nd case.

$\underset{4}{\stackrel{-}{|}}\underset{4}{\stackrel{-}{|}}\underset{3}{\stackrel{-}{|}}\underset{2}{\stackrel{\_4}{|}}$

Number of possible numbers = $4\times 4\times 3\times 2$ = 96