Key Concepts and Formulas

• Combinations: The number of ways to choose $k$ objects from a set of $n$ distinct objects, where order does not matter, is given by $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.
• Permutations with Repetition: The number of distinct arrangements of $n$ objects, where $n_1$ are of one type, $n_2$ are of another type, ..., $n_k$ are of a $k$-th type, is given by $\frac{n!}{n_1! n_2! \dots n_k!}$.
• Fundamental Principle of Counting: If there are $m$ ways to do one thing and $n$ ways to do another, then there are $m \times n$ ways to do both.

Step-by-Step Solution

Let's analyze the letters in the word BARRACK: The letters are B, A, R, R, A, C, K. Listing the distinct letters and their frequencies:

• B: 1
• A: 2
• R: 2
• C: 1
• K: 1

Total distinct letters available: 5 (B, A, R, C, K). We need to form 4-letter words. We will consider all possible cases based on the composition of the letters.

Step 1: Case 1: All four letters are distinct. In this case, we need to choose 4 different letters from the 5 distinct letters available (B, A, R, C, K).

• We use combinations to select 4 distinct letters from the 5 available distinct letters: $\binom{5}{4} = \frac{5!}{4!(5-4)!} = \frac{5!}{4!1!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{(4 \times 3 \times 2 \times 1)(1)} = 5$ This represents the number of ways to select the letters.
• For each set of 4 distinct letters, the number of ways to arrange them is $4!$. $4! = 4 \times 3 \times 2 \times 1 = 24$ This represents the number of ways to arrange the selected letters to form a word.
• The number of words for Case 1 is found by multiplying the number of ways to choose the letters by the number of ways to arrange them: $5 \times 24 = 120$

Step 2: Case 2: Two letters are identical, and the other two are distinct. For this case, we first identify which letter forms the pair (either 'A' or 'R'), and then select two distinct letters from the remaining available distinct letters.

• The letters that can form a pair are 'A' (AA) or 'R' (RR). We will consider these as subcases.

Step 3: Subcase 2a: Two 'A's and two other distinct letters.

• We fix the pair as 'A', 'A'. This has only 1 way to choose.
• We need to choose 2 distinct letters from the remaining available distinct letters. After choosing 'A', 'A', the remaining distinct letters available are B, R, C, K (4 distinct letters). We need to choose 2 from these 4: $\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)} = \frac{24}{4} = 6$
• For each selection (e.g., A, A, B, R), the number of ways to arrange them is given by the formula for permutations with repetitions (4 letters, with 'A' repeated twice): $\frac{4!}{2!} = \frac{4 \times 3 \times 2 \times 1}{2 \times 1} = \frac{24}{2} = 12$
• The number of words for Subcase 2a is found by multiplying the number of ways to choose the distinct letters by the number of ways to arrange them: $6 \times 12 = 72$

Step 4: Subcase 2b: Two 'R's and two other distinct letters.

• We fix the pair as 'R', 'R'. This has only 1 way to choose.
• We need to choose 2 distinct letters from the remaining available distinct letters. After choosing 'R', 'R', the remaining distinct letters available are B, A, C, K (4 distinct letters). We need to choose 2 from these 4: $\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)} = \frac{24}{4} = 6$
• For each selection (e.g., R, R, B, A), the number of ways to arrange them is given by the formula for permutations with repetitions (4 letters, with 'R' repeated twice): $\frac{4!}{2!} = \frac{4 \times 3 \times 2 \times 1}{2 \times 1} = \frac{24}{2} = 12$
• The number of words for Subcase 2b is found by multiplying the number of ways to choose the distinct letters by the number of ways to arrange them: $6 \times 12 = 72$

Step 5: Case 3: Two pairs of identical letters. In the word BARRACK, we have two 'A's and two 'R's. This is the only possible way to have two pairs of identical letters for a 4-letter word.

• We must choose 'A', 'A' and 'R', 'R'. There's only 1 way to choose the letters $\binom{1}{1} \times \binom{1}{1} = 1$.
• The letters are A, A, R, R. The number of ways to arrange these 4 letters, where 'A' is repeated twice and 'R' is repeated twice, is: $\frac{4!}{2!2!} = \frac{4 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)} = \frac{24}{4} = 6$
• The number of words for Case 3 is: $1 \times 6 = 6$

Step 6: Total Number of 4-letter Words To find the total number of possible 4-letter words, we sum the results from all distinct cases: Total words = (Words from Case 1) + (Words from Subcase 2a) + (Words from Subcase 2b) + (Words from Case 3) Total words = $120 + 72 + 72 + 6 = 270$

Common Mistakes & Tips

• Missing Cases: Ensure you consider all possible combinations of letter types. For example, don't forget the case where there are two pairs of identical letters.
• Double Counting: Make sure your cases are mutually exclusive. Carefully define each case to avoid overlapping.
• Selection vs. Arrangement: Remember to first select the letters using combinations and then arrange them using permutations.

Summary

To solve problems involving forming words with a given set of letters, especially when repetitions are present, it's crucial to systematically categorize all possible structures of the target word, select the required letters using combinations, arrange the selected letters using the appropriate permutation formula (accounting for repetitions if any), and sum the results from all categories to get the total number of words.

The final answer is $\boxed{270}$, which corresponds to option (D).