Conditional probability question involving balls w/o replacement A box contains 12 balls numbered 1 through 12.  If 5 balls are selected one at a time from the​ box, without​ replacement, what is the probability that the largest number selected will be 9​?
I want to just say $$\frac{9\times8\times7\times6\times5}{\binom{12}{5}}$$ but that is wrong and I don't know why.

There are $8$ choices (i.e.: 1,2,3,4,5,6,7,8) which are less than $9$, $1$ choice which equals $9$ and $3$ choices (i.e.: 10,11,12) which exceed $9$. So, appplying the hypergeometric distribution formula, we readily get: 
$$
\frac{\binom{8}{4}\binom{1}{1}\binom{3}{0}}{\binom{12}{5}}=\frac{\binom{8}{4}}{\binom{12}{5}}=\frac{\frac{5\cdot 6\cdot 7\cdot 8}{2\cdot 3\cdot 4}}{\frac{8\cdot 9\cdot 10\cdot 11\cdot 12}{2\cdot 3\cdot 4\cdot 5}}=\frac{5\cdot 6\cdot 7\cdot 8}{8\cdot 9\cdot 2\cdot 11\cdot 12}=\frac{5\cdot 7}{3\cdot 11\cdot 12}=\frac{35}{396}
$$
