Let N denote the sum of the numbers obtained when two dice are...

Given that,

${2}^{N} < N !$

$N$ is the sum of numbers of two dice.

$⇒ 2 \le N \le 12$

Let us check the given condition ${2}^{N} < N !$

$N = 1$ (not possible) $\to 0$

$N = 2$ (not possible) $\to 1$

$N = 3$ (not possible) $\to 2$

$N = 4$ (possible)

We need to find the probability for $N \ge 4$.

$⇒ P\left(N \ge 4\right) = 1 - P\left(N = 2\right) - P\left(N = 3\right)$

$= 1 - \dfrac{1}{36} - \dfrac{2}{36}$

$= \dfrac{11}{12} = \dfrac{m}{n}$

Now let us find $4m - 3n$.

$= 4 \times 11 - 3 \times 12 = 8$.

Hence, the required answer is $8$.