How many 5-digit numbers are there that include the digit 5 and exclude the digit 8? Explain your solutionit is sth about permutation , confusing...

You require five digit numbers that include 5 and exclude
8. Now the number of 5 digit numbers is 99999 - 9999 =
90000.

As the numbers need to have a 5 (though there can be
more than 1 instance) and should not have 8, we go as
follows:

Let 5 be the 1st digit. The other four digits can
have either of the other 9 choices, as we have excluded 8. This gives 9^4
numbers.

Let 5 be the 2nd digit. The first digit can have
either of 7 choices, excluding 5, 8 and 0. The other three digits can have any of the 9
choices. This gives 7 * 9^3 numbers.

Let 5 be the 3rd
digit. The first digit can have either of 7 choices, excluding 5, 8 and 0. The 2nd digit
can have 8 choices excluding 5 and 8. The other two digits can have any of the 9
choices. This gives 7 * 8 *9^2 numbers.

Let 5 be the 4th
digit. The first digit can have either of 7 choices, excluding 5, 8 and 0. The 2nd and
3rd digits can have any of 8 choices. This gives 7 * 8^2*9
numbers.

Let 5 be the 5th digit. The first digit can have
either of 7 choices, excluding 5, 8 and 0. The 2nd, 3rd and 4th digits can have any of 8
choices. This gives 7 * 8^3 numbers.

Therefore the total
possible numbers is 9^4 + 7*9^3 + 7*8*9^2 + 7*8^2*9 + 7*8^3=
23816

The required number of 5 digit numbers
is 23816.