Find the number of four letter words which can be made using the letters of the word "HIPPOPOTAMUS", if not all four letters are different.

Explanation

To find the number of four-letter words that can be formed using the letters of the word "HIPPOPOTAMUS" with the condition that not all four letters are different, we can break the problem into different cases based on the repetition of letters.

### Step-by-Step Solution:

1. **Identify the Letters and Their Frequencies:**
The word "HIPPOPOTAMUS" consists of the following letters:
- H: 1
- I: 1
- P: 3
- O: 2
- T: 1
- A: 1
- M: 1
- U: 1
- S: 1

Total unique letters = 8 (H, I, P, O, T, A, M, U, S)

2. **Total Letters:**
The total number of letters in "HIPPOPOTAMUS" is 12.

3. **Cases for Forming Four-Letter Words:**
We will consider three cases based on the repetition of letters.

**Case 1: Three letters are the same, and one letter is different.**
- The only letter that can be repeated three times is P (since it appears three times).
- The fourth letter can be any of the other letters (H, I, O, T, A, M, U, S) which gives us 8 options.
- The arrangement of the letters (PPPX, where X is the different letter) can be calculated as:
\[ \text{Total arrangements} = \frac{4!}{3!} \cdot 8 = 4 \cdot 8 = 32 \]

**Case 2: Two letters are the same and the other two letters are different.**
- The letters that can be repeated are P (3 times) and O (2 times).
- The possible combinations are:
- P, P, O, X (where X is any letter from H, I, T, A, M, U, S) gives us 7 options.
- P, P, X, Y (where X and Y are different letters from H, I, O, T, A, M, U, S) gives us \( \binom{7}{2} = 21 \) options.
- The arrangements for P, P, O, X:
\[ \text{Total arrangements} = \frac{4!}{2!} \cdot 7 = 12 \cdot 7 = 84 \]
- The arrangements for P, P, X, Y:
\[ \text{Total arrangements} = \frac{4!}{2!} \cdot 21 = 12 \cdot 21 = 252 \]
- Total for Case 2:
\[ \text{Total for Case 2} = 84 + 252 = 336 \]

**Case 3: Two letters are the same and the other two letters are the same.**
- The only combination possible is P, P, O, O.
- The arrangement for P, P, O, O is:
\[ \text{Total arrangements} = \frac{4!}{2!2!} = 6 \]

4. **Total Count:**
Now we sum the results from all cases:
\[ \text{Total} = \text{Case 1} + \text{Case 2} + \text{Case 3} = 32 + 336 + 6 = 374 \]

### Final Answer:
The total number of four-letter words that can be formed using the letters of "HIPPOPOTAMUS" with the condition that not all four letters are different is **374**.