Solution:

To get to $100$, you have to use one number which is at least $5! = 120$, because $24 \times 4 = 96$, which is less than $100$. If you use $6! = 720$ or anything larger, you need to get back from $720$ to $100$ (or further) in three seconds. Since $3 \cdot 5! < 620$, there is no way to do this in $3$ seconds. This means you have to use $5!$ at least once. The remaining numbers must get you from $120$ to $100$.

If you use three numbers all at most $3!$, you can move by at most $3 \cdot 3! = 18 < 120 - 100$. This means you have to use $4!$. From $120 - 24 = 96$, there are two ways to get to $100$: adding $6$ then subtracting $2$, or adding $2$ twice. So, to get to $100$ from $0$ in four seconds, you must either add $120$, subtract $24$, add $6$, and subtract $2$, or add $120$, subtract $24$, and add $2$ twice. You can do these steps in any order, so the first sequence yields $24$ paths and the second sequence yields $12$.

Answer: $36$