We are required to form different words with the help of letter of the word INTEGER Let `m_1` be the number of words in which I and N are never together and `m_2` be the number of words which begins with I and end with R, then `m_1//m_2` is given by

Options

A. 42
B. 30
C. 6
D. `1//30`

Answer

To solve the problem, we need to calculate \( m_1 \) and \( m_2 \) based on the word "INTEGER".

### Step 1: Calculate \( m_1 \) (Words where I and N are never together)

1. **Total Letters**: The word "INTEGER" consists of 7 letters: I, N, T, E, G, E, R.
2. **Identifying Repetitions**: The letter E is repeated twice.
3. **Total Arrangements**: The total arrangements of the letters in "INTEGER" can be calculated using the formula for permutations of multiset:
\[ \text{Total arrangements} = \frac{7!}{2!} = \frac{5040}{2} = 2520 \]

4. **Calculate arrangements where I and N are together**:
- Treat I and N as a single unit/block. Thus, we have the blocks: (IN), T, E, G, E, R, which gives us 6 blocks in total.
- The arrangements of these 6 blocks (considering E is repeated) is:
\[ \text{Arrangements with I and N together} = \frac{6!}{2!} = \frac{720}{2} = 360 \]

5. **Calculate \( m_1 \)**:
\[ m_1 = \text{Total arrangements} - \text{Arrangements with I and N together} = 2520 - 360 = 2160 \]

### Step 2: Calculate \( m_2 \) (Words that begin with I and end with R)

1. **Fix I and R**: If I is fixed at the beginning and R at the end, we have the following letters left to arrange: N, T, E, G, E (5 letters).
2. **Calculate arrangements**:
\[ m_2 = \frac{5!}{2!} = \frac{120}{2} = 60 \]

### Step 3: Calculate \( \frac{m_1}{m_2} \)

1. **Substituting values**:
\[ \frac{m_1}{m_2} = \frac{2160}{60} = 36 \]

### Final Answer:
\[ \frac{m_1}{m_2} = 36 \]