2003 AMC 12B Problem 19

The set $S$ contains $4 \cdot 4! = 96$ permutations, since the first term has $4$ choices and the remaining four terms can be arranged in $4!$ ways. For the second term to be $2,$ the first term must be $3, 4,$ or $5$ (not $1,$ not $2$), giving $3$ choices, and the remaining three terms can be arranged in $3!$ ways: $3 \cdot 3! = 18.$ The probability is $\dfrac{18}{96} = \dfrac{3}{16},$ so $a + b = 3 + 16 = 19.$ Thus, the correct answer is E.