To solve the problem of how many different four-letter words can be formed using the letters of the word "GREGARIOUS" such that each word starts with G and ends with R, we can follow these steps:
### Step 1: Identify the fixed letters
Since each word must start with G and end with R, we can denote the four-letter word as G _ _ R. Here, G is the first letter and R is the last letter.
### Step 2: Determine the available letters
The letters in "GREGARIOUS" are G, R, E, G, A, R, I, O, U, S. We have:
- G: 2
- R: 2
- E: 1
- A: 1
- I: 1
- O: 1
- U: 1
- S: 1
Since we are using one G and one R for the fixed positions, we have the following letters left to choose from for the two middle positions:
- Remaining letters: G, R, E, A, I, O, U, S (total of 8 letters)
### Step 3: Choose letters for the middle positions
We need to choose 2 letters from the remaining 8 letters (G, R, E, A, I, O, U, S). The two letters can be the same or different.
### Step 4: Calculate the arrangements
1. **Case 1: Both letters are different**
We can choose 2 different letters from the 8 available letters. The number of ways to choose 2 letters from 8 is given by the combination formula \( C(n, r) = \frac{n!}{r!(n-r)!} \):
\[
C(8, 2) = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28
\]
For each selection of 2 different letters, they can be arranged in 2! (2 factorial) ways:
\[
2! = 2
\]
Thus, the total arrangements for this case is:
\[
28 \times 2 = 56
\]
2. **Case 2: Both letters are the same**
The only letters that can be the same are G or R. So we can have:
- G G
- R R
This gives us 2 arrangements (G G and R R).
### Step 5: Combine the cases
Now, we add the results from both cases:
- From Case 1 (different letters): 56 arrangements
- From Case 2 (same letters): 2 arrangements
Total arrangements:
\[
56 + 2 = 58
\]
### Final Answer
Thus, the total number of different four-letter words that can be formed is **58**.
