Let $$N$$ denote the sum of the numbers obtained when two dice are rolled. If the probability that $$2^N < N!$$ is $$\frac{m}{n}$$, where $$m$$ and $$n$$ are coprime, $$4m - 3n$$ is equal to

1. Analyze the condition $$2^N < N!$$

For the sum $$N$$ of two dice, $$N$$ can range from $$2$$ to $$12$$.

• If $$N=2$$: $$2^2 = 4$$, $$2! = 2 \implies 4 \nless 2$$ (False)
• If $$N=3$$: $$2^3 = 8$$, $$3! = 6 \implies 8 \nless 6$$ (False)
• If $$N=4$$: $$2^4 = 16$$, $$4! = 24 \implies 16 < 24$$ (True)

The condition $$2^N < N!$$ is true for all $$N \geq 4$$.

2. Calculate the Probability

The total number of outcomes when rolling two dice is $$36$$.

We need $$P(N \geq 4)$$, which is easier to calculate as $$1 - P(N < 4)$$.

• Outcomes for $$N=2$$: $$(1,1)$$ — (1 outcome)
• Outcomes for $$N=3$$: $$(1,2), (2,1)$$ — (2 outcomes)

Total outcomes for $$N < 4$$ is $$1 + 2 = 3$$.

$$P(N \geq 4) = 1 - \frac{3}{36} = 1 - \frac{1}{12} = \frac{11}{12}$$

Here, $$m = 11$$ and $$n = 12$$ (they are coprime).

3. Final Calculation

Substitute $$m$$ and $$n$$ into the expression $$4m - 3n$$:

$$4(11) - 3(12) = 44 - 36 = \mathbf{8}$$

Correct Option: (D)