Right triangular prism $ABCDEF$ with triangular faces $\vartriangle ABC$ and $\vartriangle DEF$ and edges $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$ has $\angle ABC = 90^o$ and $\angle EAB = \angle CAB = 60^o$ . Given that $AE = 2$, the volume of $ABCDEF$ can be written in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Compute $m + n$.
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1. **Identify the given information and the shape of the prism:**
   - The prism is a right triangular prism with triangular faces $\vartriangle ABC$ and $\vartriangle DEF$.
   - $\angle ABC = 90^\circ$ and $\angle EAB = \angle CAB = 60^\circ$.
   - $AE = 2$.

2. **Determine the dimensions of the triangular base $\vartriangle ABC$:**
   - Since $\angle EAB = \angle CAB = 60^\circ$, $\vartriangle ABE$ is an equilateral triangle with side length $AE = 2$.
   - Therefore, $AB = BE = 2$.
   - Since $\angle ABC = 90^\circ$, $\vartriangle ABC$ is a 30-60-90 triangle.
   - In a 30-60-90 triangle, the sides are in the ratio $1 : \sqrt{3} : 2$.
   - Thus, $BC = AB \cdot \sqrt{3} = 2 \cdot \sqrt{3} = 2\sqrt{3}$.

3. **Calculate the area of the triangular base $\vartriangle ABC$:**
   - The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
   - Here, the base $AB = 2$ and the height $BC = 2\sqrt{3}$.
   - Therefore, the area of $\vartriangle ABC$ is:
     \[
     \text{Area of } \vartriangle ABC = \frac{1}{2} \times 2 \times 2\sqrt{3} = 2\sqrt{3}
     \]

4. **Determine the height of the prism:**
   - The height of the prism is the distance between the triangular bases, which is $AD = BE = CF$.
   - Since $\vartriangle ABE$ is an equilateral triangle with side length $2$, the height of the prism is $2$.

5. **Calculate the volume of the prism:**
   - The volume of a prism is given by the area of the base times the height.
   - Therefore, the volume of the prism is:
     \[
     \text{Volume} = \text{Area of base} \times \text{height} = 2\sqrt{3} \times 2 = 4\sqrt{3}
     \]

6. **Express the volume in the form $\frac{m}{n}$:**
   - The volume $4\sqrt{3}$ can be written as $\frac{4\sqrt{3}}{1}$.
   - Here, $m = 4\sqrt{3}$ and $n = 1$.

7. **Compute $m + n$:**
   - Since $m = 4\sqrt{3}$ and $n = 1$, we have:
     \[
     m + n = 4\sqrt{3} + 1
     \]

The final answer is $\boxed{5}$.