How many words can be formed using the letters of the word ASSESSMENT if each word begin with A and end with T?

Answer

To solve the problem of how many words can be formed using the letters of the word "ASSESSMENT" such that each word begins with 'A' and ends with 'T', we will follow these steps:

### Step 1: Identify the letters in the word "ASSESSMENT"
The word "ASSESSMENT" consists of the following letters:
- A: 1
- S: 4
- E: 2
- M: 1
- N: 1
- T: 1

In total, there are 11 letters.

### Step 2: Fix the positions of 'A' and 'T'
Since each word must start with 'A' and end with 'T', we can fix 'A' at the beginning and 'T' at the end. This means we only need to arrange the letters in between.

### Step 3: Count the remaining letters
After fixing 'A' and 'T', we are left with the following letters to arrange:
- S: 4
- E: 2
- M: 1
- N: 1

This gives us a total of 8 letters (4 S's, 2 E's, 1 M, and 1 N).

### Step 4: Calculate the arrangements of the remaining letters
To find the number of distinct arrangements of these 8 letters, we use the formula for permutations of multiset:

\[ \text{Number of arrangements} = \frac{n!}{n_1! \times n_2! \times n_3! \times \ldots} \]

Where:
- \( n \) is the total number of items to arrange,
- \( n_1, n_2, \ldots \) are the counts of each distinct item.

In our case:
- Total letters to arrange (n) = 8
- Count of S's (n1) = 4
- Count of E's (n2) = 2
- Count of M's (n3) = 1
- Count of N's (n4) = 1

So the formula becomes:

\[ \text{Number of arrangements} = \frac{8!}{4! \times 2! \times 1! \times 1!} \]

### Step 5: Calculate the factorials
Now we calculate the factorials:
- \( 8! = 40320 \)
- \( 4! = 24 \)
- \( 2! = 2 \)
- \( 1! = 1 \)

### Step 6: Substitute the values into the formula
Now substituting the values into our formula:

\[ \text{Number of arrangements} = \frac{40320}{24 \times 2 \times 1 \times 1} = \frac{40320}{48} = 840 \]

### Conclusion
Thus, the total number of distinct words that can be formed using the letters of "ASSESSMENT" that begin with 'A' and end with 'T' is **840**.