Key Concepts and Formulas

• Combinations: The number of ways to choose $r$ objects from a set of $n$ distinct objects is given by $^nC_r = \frac{n!}{r!(n-r)!}$.
• Permutations with Repetition: The number of permutations of $n$ objects, where $n_1$ are alike, $n_2$ are alike, ..., $n_k$ are alike, is given by $\frac{n!}{n_1! n_2! ... n_k!}$.

Step-by-Step Solution

Step 1: Analyze the Word and Letter Frequencies

We are given the word 'DISTRIBUTION'. We need to find the number of four-letter words that can be formed using the letters of this word. First, we find the frequency of each letter:

• D: 1
• I: 3
• S: 1
• T: 2
• R: 1
• B: 1
• U: 1
• O: 1
• N: 1

There are 9 distinct letters: D, I, S, T, R, B, U, O, N. 'I' appears 3 times, and 'T' appears 2 times.

Step 2: Categorize Selections of Four Letters

We need to consider all possible distinct patterns of repetition for a group of four letters. The possible categories are:

1. All four letters are distinct.
2. Two letters are alike, and the other two are distinct.
3. Two letters are alike of one kind, and two letters are alike of another kind.
4. Three letters are alike, and one is distinct.

Step 3: Calculate Permutations for Each Category

Case 1: All four letters are distinct (a, b, c, d)

• Sub-step 3.1: Choose the letters (Combinations) We have 9 distinct letters, and we need to choose 4. The number of ways to do this is $^9C_4$. $^9C_4 = \frac{9!}{4!5!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$

• Sub-step 3.2: Arrange the chosen letters (Permutations) For each selection of 4 distinct letters, we can arrange them in $4!$ ways. $4! = 4 \times 3 \times 2 \times 1 = 24$

• Sub-step 3.3: Total permutations for this case Total permutations for this case is $126 \times 24 = 3024$.

Case 2: Two letters are alike, and the other two are distinct (a, a, b, c)

• Sub-step 3.1: Choose the letters (Combinations) First, we choose the letter that appears twice. Only 'I' and 'T' can be chosen. So, there are $^2C_1 = 2$ ways to choose the repeated letter. Then, we choose two distinct letters from the remaining 8 distinct letters. This can be done in $^8C_2$ ways. $^8C_2 = \frac{8!}{2!6!} = \frac{8 \times 7}{2 \times 1} = 28$ Total number of ways to select the letters is $2 \times 28 = 56$.

• Sub-step 3.2: Arrange the chosen letters (Permutations) For each selection, we have 4 letters where one letter is repeated twice. The number of arrangements is $\frac{4!}{2!}$. $\frac{4!}{2!} = \frac{24}{2} = 12$

• Sub-step 3.3: Total permutations for this case Total permutations for this case is $56 \times 12 = 672$.

Case 3: Two letters are alike of one kind, and two letters are alike of another kind (a, a, b, b)

• Sub-step 3.1: Choose the letters (Combinations) We need to select two letters that appear at least twice. Only 'I' and 'T' satisfy this condition. So we choose both 'I' and 'T'. Number of ways to choose the letters is 1.

• Sub-step 3.2: Arrange the chosen letters (Permutations) For the selection {I, I, T, T}, the number of arrangements is $\frac{4!}{2!2!}$. $\frac{4!}{2!2!} = \frac{24}{4} = 6$

• Sub-step 3.3: Total permutations for this case Total permutations for this case is $1 \times 6 = 6$.

Case 4: Three letters are alike, and one is distinct (a, a, a, b)

• Sub-step 3.1: Choose the letters (Combinations) The only letter that appears three times is 'I'. So, we must choose 'I' for the repeated letter. Then we choose one letter from the remaining 8 distinct letters (D, S, T, R, B, U, O, N). This can be done in $^8C_1$ ways. $^8C_1 = 8$

• Sub-step 3.2: Arrange the chosen letters (Permutations) For each selection, we have 4 letters where one letter is repeated three times. The number of arrangements is $\frac{4!}{3!}$. $\frac{4!}{3!} = \frac{24}{6} = 4$

• Sub-step 3.3: Total permutations for this case Total permutations for this case is $8 \times 4 = 32$.

Step 4: Sum the Permutations

The total number of words is the sum of the permutations from all four cases: $3024 + 672 + 6 + 32 = 3734$

Common Mistakes & Tips

• Remember to consider all possible cases of letter repetition.
• Carefully calculate combinations and permutations, paying attention to repeated letters.
• It is helpful to list the letters and their frequencies at the beginning.

Summary

We analyzed the word 'DISTRIBUTION' and its letter frequencies. We categorized the possible selections of four letters based on repetition and calculated the number of permutations for each category. Finally, we summed the permutations from all cases to obtain the total number of four-letter words that can be formed.

The final answer is \boxed{3734}.