1. Key Concepts and Formulas

• Probability of an Event: For an event $E$ within a sample space $S$ of equally likely outcomes, its probability $P(E)$ is given by: $P(E) = \frac{\text{Number of outcomes favorable to } E}{\text{Total number of possible outcomes in } S}$
• Sample Space for Rolling Two Dice: When two standard six-sided dice are rolled, the total number of possible outcomes is $6 \times 6 = 36$. Each outcome is an ordered pair (Die 1 result, Die 2 result).
• Complementary Probability: If $E$ is an event, its complement $E'$ is the event that $E$ does not occur. The probability of $E$ can be found using the formula: $P(E) = 1 - P(E')$ This is especially useful when the number of outcomes for $E'$ is significantly smaller or easier to determine than for $E$.

2. Step-by-Step Solution

Step 1: Determine the Sample Space for N (Sum of Two Dice) and its Frequencies. The variable $N$ is the sum of the numbers obtained when two dice are rolled. We need to list all possible values of $N$ and the number of ways each sum can occur. This forms the basis for calculating probabilities.

The total number of possible outcomes when rolling two dice is $6 \times 6 = 36$.

Step 2: Analyze the Inequality $2^N < N!$. We need to find for which values of $N$ (from 2 to 12) the inequality $2^N < N!$ holds true.

• For $N=2$:

  • $2^N = 2^2 = 4$
  • $N! = 2! = 2 \times 1 = 2$
  • Is $4 < 2$? No. So, $N=2$ does not satisfy the inequality. ($2^N \ge N!$ holds)

• For $N=3$:

  • $2^N = 2^3 = 8$
  • $N! = 3! = 3 \times 2 \times 1 = 6$
  • Is $8 < 6$? No. So, $N=3$ does not satisfy the inequality. ($2^N \ge N!$ holds)

• For $N=4$:

  • $2^N = 2^4 = 16$
  • $N! = 4! = 4 \times 3 \times 2 \times 1 = 24$
  • Is $16 < 24$? Yes. So, $N=4$ satisfies the inequality.

• For $N=5$:

  • $2^N = 2^5 = 32$
  • $N! = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$
  • Is $32 < 120$? Yes. So, $N=5$ satisfies the inequality.

Reasoning for $N \ge 4$: Once $N!$ becomes greater than $2^N$ (which happens at $N=4$), it will continue to be greater for all subsequent integer values of $N$. This is because when $N$ increases to $N+1$:

• $2^{N+1} = 2^N \times 2$
• $(N+1)! = N! \times (N+1)$ Since for $N \ge 4$, $(N+1) > 2$, the factorial term $(N+1)!$ grows much faster than the exponential term $2^{N+1}$. Therefore, the inequality $2^N < N!$ holds for all values of $N$ such that $N \ge 4$. The values of $N$ that satisfy $2^N < N!$ are $\{4, 5, 6, 7, 8, 9, 10, 11, 12\}$. The values of $N$ that satisfy $2^N \ge N!$ are $\{2, 3\}$.

Step 3: Calculate the Required Probability using Complementary Events. Let $E$ be the event that $2^N < N!$. The values of $N$ for which $E$ occurs are $\{4, 5, 6, 7, 8, 9, 10, 11, 12\}$. There are 9 such values. The complementary event $E'$ is $2^N \ge N!$. The values of $N$ for which $E'$ occurs are $\{2, 3\}$. There are only 2 such values, making it much simpler to calculate $P(E')$.

• Calculate the number of favorable outcomes for $E'$ (i.e., $N=2$ or $N=3$):

  • From Step 1, the frequency for $N=2$ is 1.
  • From Step 1, the frequency for $N=3$ is 2.
  • Since these are mutually exclusive, the total number of ways for $E'$ is $1 + 2 = 3$.

• Calculate the probability of the complementary event $P(E')$:

  • Total number of possible outcomes is 36.
  • $P(E') = \frac{\text{Number of ways for } N=2 \text{ or } N=3}{\text{Total number of outcomes}} = \frac{3}{36} = \frac{1}{12}$

• Calculate the probability of the event $E$ using complementary probability:

  • $P(E) = 1 - P(E')$
  • $P(E) = 1 - \frac{1}{12} = \frac{12 - 1}{12} = \frac{11}{12}$ So, the probability that $2^N < N!$ is $\frac{11}{12}$.

Step 4: Calculate $4m-3n$. The problem states that the probability is $\frac{m}{n}$, where $m$ and $n$ are coprime. From Step 3, we found the probability to be $\frac{11}{12}$. Thus, we have $m=11$ and $n=12$. Let's check if $m$ and $n$ are coprime: The greatest common divisor of 11 and 12 is 1, so they are indeed coprime.

Now, we substitute these values into the expression $4m-3n$: $4m - 3n = 4(11) - 3(12)$ $4m - 3n = 44 - 36$ $4m - 3n = 8$

3. Common Mistakes & Tips

• Incomplete Sample Space: Ensure all 36 possible outcomes for two dice rolls are considered, and the frequencies for each sum $N$ are correctly tabulated. Remember that (1,2) and (2,1) are distinct outcomes.
• Errors in Inequality Evaluation: Carefully calculate $2^N$ and $N!$ for each $N$. A common mistake is to miscalculate factorials or powers.
• Missing the Trend: Failing to observe that $N!$ grows significantly faster than $2^N$ for $N \ge 4$ could lead to unnecessary calculations for all $N$ from 6 to 12. Recognizing this pattern is a key time-saver.
• Not Using Complementary Probability: Directly summing frequencies for $N \in \{4, \dots, 12\}$ is more prone to error and takes longer than calculating for $N \in \{2, 3\}$ and using the complementary rule.
• Simplifying Fractions: Always reduce the probability fraction to its simplest form to ensure $m$ and $n$ are coprime before the final calculation.

4. Summary

This problem required a systematic approach combining probability fundamentals with number theory. We first established the sample space for the sum of two dice rolls and their frequencies. Next, we analyzed the inequality $2^N < N!$ for various values of $N$, identifying the specific values for which it holds. By recognizing the rapid growth of the factorial function, we efficiently determined the set of $N$ values satisfying the inequality. Finally, employing the principle of complementary probability significantly simplified the calculation of the required probability, leading to $P(E) = \frac{11}{12}$. With $m=11$ and $n=12$, the final expression $4m-3n$ was computed as $8$.

5. Final Answer

The final answer is $\boxed{8}$, which corresponds to option (C).