Question 1177880


 if repetitions are allowed,

Two ways to get the answer:

1.  There are{{{ 9999}}} integers starting with {{{1 }}}and ending with {{{9999}}}.  
But the {{{999}}} integers starting with {{{1 }}}and ending with {{{999}}} have less than {{{4 }}}digits,

so the desired number is {{{9999-999}}} or {{{9000}}} ways.

2.
there is total of {{{10}}} digits from {{{0}}} to {{{9}}}
so, there is {{{10}}} ways to choose each ones, tens, hundredths, and thousands'
that is {{{10^4}}}
since {{{0 }}}cannot be in place of thousands', only ones, tens, hundredths which is  {{{10^3}}} three digit numbers , deduct it

{{{10^4-10^3=10000-1000=9000}}}