Question

how many five digit numbers include the digits 4 or 6 or both?

Answer 1

answer)

First, we will find out the total number of five-digit numbers.

we know that there is a total of 10 digits (0,1,2,3,4,5,6,7,8,9)

and to have a five digit number, the first digit should not be = 0 ( as in that case the number would be of 4 digits if first place would be 0)

so the number of possible digits for first place is = 10-1 = 9

and for the rest of the places, we can have 0 so number of possible digits for the rest of the places is = 10

so total number of possible numbers with 5 digits = 9*10*10*10*10

= 90000

now the total number of 5 digit numbers with no 4 and no 6

as we don't want 4 and 6

possible digits for first place are 1,2,3,5,7,8,9 that is 7

and for the rest of the places we have, 0,1,2,3,5,7,8,9 that is 8

so, the total number of 5 digit numbers with no 4 and no 6 = 7*8*8*8*8 = 28672

now if we will subtract the total number of 5 digit numbers with no 4 and no 6 from the total number of five-digit numbers.

we will get the five digit numbers with the digits 4 or 6 or both

final answer

= 90000-28672

= 61328