There are 10 single-digit numbers.

For larger numbers of digits, there's only 9 choices for the first one, as zero's excluded there. There are still 9 choices for the second digits, as zero has now appeared, but the one chosen for first is no longer available. Each subsequent digit has one fewer choices:

10

9*9=81

9*9*8=648

9*9*8*7=4536

9*9*8*7*6=27216

9*9*8*7*6*5=136080

9*9*8*7*6*5*4=544320

9*9*8*7*6*5*4*3=1632960

9*9*8*7*6*5*4*3*2=3265920

9*9*8*7*6*5*4*3*2*1=3265920

tot =

     8877691

     

including the initial 10 1-digit numbers.

The last two terms are the same, as one determines the digit that's not present and the other determines the digit that's last.

To print the above table:

tot=10;

s='9*9';

for i=8:-1:0

  disp([s '=' num2str(eval(s))])

  tot=tot+eval(s);

  s=[s '*' num2str(i)];

end

tot