Question 852124
Yes and no. 
I think you've got the concepts but you're missing a lot of terms. 
Since you don't know when the ace is served, the example you gave is only one possible outcome. 
It could have been 
AAANNNNN
or 
ANAANNNN
or
NNNNNAAA
So you need to capture all of the possible outcomes that each individually have a probability of {{{3125/6^8}}}
You can use Pascal's triangle with 9 terms (1,8,28,56,70,56,28,8,1).
So for no aces, there is 1 term (NNNNNNNN)
For 1 ace, 8 terms (ANNNNNNN, NANNNNNN, .... , NNNNNNNA)
For 2 aces, 28 terms (too many terms to write out)
For 3 aces, 56 terms.
So then,

{{{highlight(P(3)=56*(1/6)^3*(5/6)^5=0.10419)}}}
The other terms are,
{{{P(0)=1*(1/6)^0*(5/6)^8=0.23257}}}
{{{P(1)=8*(1/6)^1*(5/6)^7=0.37211}}}
{{{P(2)=28*(1/6)^2*(5/6)^6=0.26048}}}
{{{P(4)=70*(1/6)^4*(5/6)^4=0.02605}}}
{{{P(5)=56*(1/6)^5*(5/6)^3=0.00417}}}
{{{P(6)=28*(1/6)^6*(5/6)^2=0.00042}}}
{{{P(7)=8*(1/6)^7*(5/6)^1 =0.000024}}}
{{{P(8)=1*(1/6)^8*(5/6)^0 =0.0000006}}}

If you add all the terms, they do sum to 1.