In sub - parts (i) to (x) choose the correct option and in sub - parts (xi) to (xv) , answer the questions an instructed.
How many words , with or without meaning ,can be formed using the letters of the word ENGINEERING ?

Answer

To find the number of words that can be formed using the letters of the word "ENGINEERING", we will follow these steps:

### Step 1: Identify the total number of letters and their frequencies
The word "ENGINEERING" consists of 11 letters in total. The frequency of each letter is as follows:
- E: 3
- N: 3
- G: 2
- I: 2
- R: 1

### Step 2: Use the formula for permutations of multiset
The formula to calculate the number of distinct permutations of a multiset is given by:

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

Where:
- \(N\) is the total number of items,
- \(n_1, n_2, n_3, \ldots\) are the frequencies of the identical items.

### Step 3: Substitute the values into the formula
In our case:
- \(N = 11\) (total letters)
- \(n_1 = 3\) (for E)
- \(n_2 = 3\) (for N)
- \(n_3 = 2\) (for G)
- \(n_4 = 2\) (for I)
- \(n_5 = 1\) (for R)

Thus, we can write:

\[ \text{Number of permutations} = \frac{11!}{3! \times 3! \times 2! \times 2! \times 1!} \]

### Step 4: Calculate the factorial values
Now we will calculate the factorials:
- \(11! = 39916800\)
- \(3! = 6\)
- \(2! = 2\)
- \(1! = 1\)

### Step 5: Substitute the factorial values into the equation
Now substituting the factorial values into our formula:

\[ \text{Number of permutations} = \frac{39916800}{6 \times 6 \times 2 \times 2 \times 1} \]

Calculating the denominator:

\[ 6 \times 6 = 36 \]
\[ 36 \times 2 = 72 \]
\[ 72 \times 2 = 144 \]

So, we have:

\[ \text{Number of permutations} = \frac{39916800}{144} \]

### Step 6: Perform the division
Now we perform the division:

\[ 39916800 \div 144 = 277200 \]

### Final Answer
Thus, the total number of words that can be formed using the letters of the word "ENGINEERING" is **277200**.

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