Let $S$ be the [set](https://artofproblemsolving.com/wiki/index.php/Set) of [permutations](https://artofproblemsolving.com/wiki/index.php/Permutation) of the [sequence](https://artofproblemsolving.com/wiki/index.php/Sequence) $1,2,3,4,5$ for which the first term is not $1$. A permutation is chosen randomly from $S$. The [probability](https://artofproblemsolving.com/wiki/index.php/Probability) that the second term is $2$, in lowest terms, is $a/b$. What is $a+b$? 
$\mathrm{(A)}\ 5 \qquad\mathrm{(B)}\ 6 \qquad\mathrm{(C)}\ 11 \qquad\mathrm{(D)}\ 16 \qquad\mathrm{(E)}\ 19$
There are $4$ choices for the first element of $S$, and for each of these choices there are $4!$ ways to arrange the remaining elements. If the second element must be $2$, then there are only $3$ choices for the first element and $3!$ ways to arrange the remaining elements. Hence the answer is $\frac{3 \cdot 3!}{4 \cdot 4!} = \frac {18}{96} = \frac{3}{16}$, and $a+b=19 \Rightarrow \mathrm{(E)}$.