If eight different books, 5 math and 3 cs, are randomly arranged on a shelf, what's the probability that all 3 cs books are together? Here's what I did, but I feel like it's wrong. 
$$3C3 * 5!/8!$$
so that we're choosing all 3 cs books to be together, and then choosing the remaining 5 math books to be arranged on the shelf, and we divide by $8!$ since there are $8!$ ways of arranging all the books. 
 A: Your answer is wrong because there are not $\binom 3 3$ possible locations for the CS books. You should see that as relatively obvious since $\binom 3 3 = 1$ which obviously doesn't make sense.

Let's try a simpler approach. Imagine treating the $3$ CS books as a single book in the group of $8$ - which becomes $6$ books in light of this interpretation. By treating the three as a single book, this guarantees the three are together, making things much simpler!
How many ways can you arrange this group of $6$ books? That would be $6!$ ways.
Within the triplet of three books, you can arrange them in $3!$ further ways. We do care about the order, which is why we don't use combinations here. Since this lends us to having $3!$ permutations per permutation of the overall set with the triplet, we multiply $6!$ and $3!$.
Overall, there are $8!$ ways to arrange the set of books altogether. Thus, the probability the three end up together is given by
$$\frac{\text{desired arrangements}}{\text{all possible arrangements}} = \frac{6! \cdot 3!}{8!} = \frac{3}{28} \approx 10.714\%$$
A: As an alternative to the nice solution provided by Eevee Trainer, we take as our sample space the $\binom{8}{3}$ ways of selecting three of the eight positions for the computer science books.  If the three computer science books are placed consecutively, the leftmost book in the block of three computer science books must be placed in one of the first six positions.  Hence, the probability that the three computer science books are placed consecutively is
$$\frac{6}{\binom{8}{3}} = \frac{3}{28}$$
