smc

The probability could be written as a fraction where $\frac{\text{number of viable solutions in S}}{\text{total permutations in S}}$. The total number of permutations is essentially $5!$, and the number of permutations where $1$ is the first number is $4!$, therefore the number of permutations in $S$ is $5!-4!$. For the first number in the desired solution, there are 3 options ($3$, $4$, $5$). For the second number in the desired solution, there is only one option ($2$). For the third number in the desired solution, there are 3 options ($1$, and the numbers not used in the first digit). For the fourth number in the desired solution, there are 2 options (numbers not used in the third digit). For the last number in the desired solution, there is only one option (number not used in the fourth digit). Therefore, $3\cdot 1\cdot 3\cdot 2\cdot 1$ will be the number of desired outcomes. Finally, $\frac{3\cdot 1\cdot 3\cdot 2\cdot 1}{5!-4!}$ equates to $\frac{18}{96}$, which is $\frac{3}{16}$. The answer is then $3+16=\boxed{19}$.