To find the number of permutations of 4 letters from the word "ARRANGEMENT", we first need to analyze the letters and their frequencies. The letters in "ARRANGEMENT" are as follows:
- A: 2
- R: 2
- N: 2
- G: 1
- E: 2
- M: 1
- T: 1
This gives us a total of 11 letters, but since some letters are repeated, we will consider different cases based on the frequency of letters when forming 4-letter combinations.
### Step 1: Identify Cases for Permutations
We can break down the problem into different cases based on the frequency of letters:
1. **Case 1: All four letters are different.**
2. **Case 2: Two letters are the same, and the other two are different.**
3. **Case 3: Two pairs of letters are the same.**
4. **Case 4: Three letters are the same (not possible here).**
5. **Case 5: Four letters are the same (not possible here).**
### Step 2: Calculate Each Case
#### Case 1: All four letters are different
We have 7 different letters: A, R, N, G, E, M, T. We need to choose 4 out of these 7 letters.
The number of ways to choose 4 letters from 7 is given by \( \binom{7}{4} \), and since all letters are different, we can arrange these 4 letters in \( 4! \) ways.
\[
\text{Number of ways} = \binom{7}{4} \times 4! = 35 \times 24 = 840
\]
#### Case 2: Two letters are the same, and the other two are different
We can choose one letter that appears twice (A, R, N, or E) and then choose 2 different letters from the remaining letters.
- Choose 1 letter from {A, R, N, E} (4 choices).
- Choose 2 letters from the remaining 6 letters.
The number of ways to choose 2 letters from 6 is \( \binom{6}{2} \), and these can be arranged in \( \frac{4!}{2!} \) ways (since two letters are the same).
\[
\text{Number of ways} = 4 \times \binom{6}{2} \times \frac{4!}{2!} = 4 \times 15 \times 12 = 720
\]
#### Case 3: Two pairs of letters are the same
We can choose 2 letters that appear twice (A, R, N, E). The number of ways to choose 2 letters from these 4 is \( \binom{4}{2} \), and these can be arranged in \( \frac{4!}{2! \times 2!} \) ways.
\[
\text{Number of ways} = \binom{4}{2} \times \frac{4!}{2! \times 2!} = 6 \times 6 = 36
\]
### Step 3: Combine All Cases
Now, we sum the results from all the cases:
\[
\text{Total permutations} = 840 + 720 + 36 = 1596
\]
### Final Answer
Thus, the total number of permutations of 4 letters from the word "ARRANGEMENT" is **1596**.
