Odds of winning a superlottery The problem:
There's a lottery (or a "superlottery", I'm not really sure how that's different). In order to play, I select 8 numbers from the first 90 positive integers (so, 1-90 inclusive). Also, a computer selects 12 numbers from the first 90 positive integers. If all of my 8 numbers are in the set of 12 selected by the computer, I win. Assuming all numbers are randomly selected, what are the odds of winning?
My solution:
Once the computer has chosen its 12 numbers, there are $\binom{12}{8}$ ways for me to pick a winning set of numbers, and $\binom{90}{8}$ ways to pick a set of 8 numbers overall. So, the odds of winning are $\frac{\binom{12}{8}}{\binom{90}{8}}=\frac{1}{156597013}\approx6.38*10^{-9}$.
My friend's solution:
There are $\binom{12}{8}$ ways to choose winning numbers. The odds of the first number matching are $\frac{12}{90}$, the odds of the second number matching are $\frac{11}{89}$, and so on. So the odds of all the numbers matching are $\binom{12}{8}*\frac{12}{90}*...*\frac{5}{83}=\frac{495}{156597013}\approx2.107*10^{-6}$.
Which of us is right (if either), and what mistake is the other person making?

Your solution is correct and your friend made a common mistake.
Suppose the winning numbers are $\{1,2,3,4,5,6,7,8,9,10,11,12\}$. 
The first number selected must be in this set with probability $\frac{12}{90}$.
The second number selected must be one of the remaining numbers with probability $\frac{11}{89}$.
and so forth giving
$$\begin{align*}
p=\frac{12}{90}\cdot\frac{11}{89}\cdot\frac{10}{88}\cdot\frac{9}{87}\cdot\frac{8}{86}\cdot\frac{7}{85}\cdot\frac{6}{84}\cdot\frac{5}{83}
&\approx6.386\cdot10^{-9}
\end{align*}$$
which agrees with your solution.
Using this method, there is no need to account for how the numbers are arranged since we arbitrarily let the first number selected to be any of the $12$, the second number selected to be any of the remaining $11$, etc.
Note: 
It's easy to show that
$$\frac{12}{90}\cdot\frac{11}{89}\cdot\frac{10}{88}\cdot\frac{9}{87}\cdot\frac{8}{86}\cdot\frac{7}{85}\cdot\frac{6}{84}\cdot\frac{5}{83}=\frac{12 \choose 8}{90 \choose 8}$$
We have that
$$\begin{align*}
\frac{12 \choose 8}{90 \choose 8}
&=\frac{\frac{12!}{8!4!}}{\frac{90!}{82!8!}}\\\\
&=\frac{12\cdot11\cdot10\cdot9}{\color{red}{{4\cdot3\cdot2\cdot1}}}\cdot\frac{8\cdot7\cdot6\cdot5\cdot\color{red}{4\cdot3\cdot2\cdot1}}{90\cdot89\cdot88\cdot87\cdot86\cdot85\cdot84\cdot83}\\\\
&=\frac{12}{90}\cdot\frac{11}{89}\cdot\frac{10}{88}\cdot\frac{9}{87}\cdot\frac{8}{86}\cdot\frac{7}{85}\cdot\frac{6}{84}\cdot\frac{5}{83}
\end{align*}$$
