Question 1050878
there are {{{10}}} digits from {{{0}}} to {{{9}}}

when {{{0}}} and {{{7}}} cannot be the leading number, then there is {{{8}}} choices for the {{{first}}} digit and they are: {{{1}}}, {{{2}}},{{{ 3}}}, {{{4}}}, {{{5}}}, {{{6}}}, {{{8}}}, and {{{9}}}

there will be {{{10}}} choices for {{{each}}} of the remaining {{{6}}} digits, and we have:

{{{8*10^6 = 8000000}}}

or, you can do it this way:

Position: {{{1}}}st.. {{{2}}}nd ..{{{3}}}rd.. {{{4}}}th.. {{{5}}}th.. {{{6}}}th ..{{{7}}}th 
choices....{{{8 }}}.... {{{10}}}......{{{ 10}}}..... {{{ 10 }}}......{{{ 10}}}....{{{ 10}}}..... {{{10}}} 

total number of choices: {{{8 * 10 * 10 * 10 * 10 * 10 * 10 =8*1000000=8000000}}}