Question: A linguist analyzing word evolution models a simplified language where each word is a sequence of 8 characters, each chosen independently from the set $\{a, b, c\}$. What is the probability that a randomly generated word contains exactly 3 $a$’s, 2 $b$’s, and 3 $c$’s?

["Title: Probability of Specific Character Composition in a Constructed 8-ACharacter Language Model", "In the realm of linguistic modeling, rebuilding language structures through combinatorics helps uncover underlying patterns in word formation. A recent study models a simplified synthetic language where every word consists of exactly 8 characters, each selected independently and uniformly from the alphabet ${a, b, c}$. This setup offers a probabilistic framework to analyze rare character distributions—such as a word containing exactly 3 $a$s, 2 $b$s, and 3 $c$s.", "Let’s explore how to calculate the probability of generating such a word, reflecting real principles in combinatorics and probabilistic linguistics.", "---", "### Understanding the Model", "Each position in the word is independently chosen from ${a, b, c}$ with equal probability:\n$$\nP(a) = P(b) = P(c) = \frac{1}{3}\n$$\nThe total length of each word is fixed at $n = 8$, and we seek the probability that a randomly generated word contains:\n- Exactly $n_a = 3$ occurrences of $a$,\n- Exactly $n_b = 2$ occurrences of $b$,\n- Exactly $n_c = 3$ occurrences of $c$.", "Note: $3 + 2 + 3 = 8$, so this distribution is valid within the word length.", "---", "### Application of Multinomial Distribution", "The number of such words follows a multinomial distribution. The probability mass function for this scenario is:", "$$\nP(n_a = 3, n_b = 2, n_c = 3) = \frac{8!}{3!,2!,3!} \left(\frac{1}{3}\right)^3 \left(\frac{1}{3}\right)^2 \left(\frac{1}{3}\right)^3\n$$", "Since the probabilities for each character are equal, the expression simplifies to:", "$$\nP = \frac{8!}{3!,2!,3!} \cdot \left(\frac{1}{3}\right)^8\n$$", "---", "### Step-by-Step Computation", "First, compute the multinomial coefficient:", "$$\n\frac{8!}{3!,2!,3!} = \frac{40320}{6 \cdot 2 \cdot 6} = \frac{40320}{72} = 560\n$$", "Next, compute the probability factor:", "$$\n\left(\frac{1}{3}\right)^8 = \frac{1}{6561}\n$$", "Now combine both:", "$$\nP = 560 \cdot \frac{1}{6561} = \frac{560}{6561}\n$$", "This fraction is already in simplest form (560 and 6561 share no common factors other than 1).", "---", "### Interpretation and Significance", "This result quantifies the chance of a precisely balanced character distribution in a synthetically controlled language model. Such models help linguists simulate and predict how structure might emerge in natural languages from simple stochastic rules.", "For anyone interested in computational linguistics or language evolution, this example illustrates how combinatorics underpins probabilistic assumptions in word generation. Whether modeling ancestral language splits or artificial dialects, understanding exact character frequencies strengthens predictive frameworks.", "---", "### Final Answer", "The probability that a randomly generated 8-character word in this model contains exactly 3 $a$s, 2 $b$s, and 3 $c$s is:", "$$\n\boxed{\frac{560}{6561}}\n$$"]