Let $N$ be the sum when two fair dice are rolled.

The possible values of $N$ are: $2,3,4,\dots,12$

We need the event: $2^N < N!$

Let us check for each possible $N$ from $2$ onward.

Compare $2^N$ and $N!$:

• For $N=2$: $2^2=4,\quad 2!=2$ So $4<2$ is false.

• For $N=3$: $2^3=8,\quad 3!=6$ So $8<6$ is false.

• For $N=4$: $2^4=16,\quad 4!=24$ So $16<24$ is true.

Now for all $N\ge 4$, factorial grows faster than powers of $2$, so the inequality remains true.

Hence the favorable values are: $N=4,5,6,7,8,9,10,11,12$

Total number of outcomes when two dice are rolled: $6\times 6=36$

Unfavorable sums are only $N=2$ and $N=3$.

• Sum $2$: $(1,1)$ → $1$ outcome
• Sum $3$: $(1,2),(2,1)$ → $2$ outcomes

So unfavorable outcomes $=1+2=3$.

Therefore favorable outcomes: $36-3=33$

Hence the required probability is: $\frac{33}{36}=\frac{11}{12}$

So, $m=11,\quad n=12$

Compute: $4m-3n=4(11)-3(12)=44-36=8$

Therefore the correct option is: $\boxed{8}$ which is option $\boxed{\text{C}}$.

Comparison with stored answer:

Stored correct answer is $\text{C}$, which corresponds to $8$. This matches our derived answer.