Letters of the word INDIANOIL are arranged in all possible ways. The number of permutations in which A, I, O occur only at odd places, is:

Options

A. 720
B. 360
C. 240
D. 120

Answer

To solve the problem of arranging the letters of the word "INDIANOIL" such that the letters A, I, and O occur only at odd places, we will follow these steps:

### Step 1: Identify the Odd Positions
The word "INDIANOIL" has 9 letters. The odd positions in a 9-letter word are:
1. Position 1
2. Position 3
3. Position 5
4. Position 7
5. Position 9

### Step 2: Choose Positions for A, I, and O
We need to place the letters A, I, and O in 3 of these 5 odd positions. The remaining 2 odd positions will be filled with the other letters.

### Step 3: Calculate the Ways to Choose Positions
We can choose 3 positions out of 5 for A, I, and O. The number of ways to choose 3 positions from 5 is given by the combination formula:
\[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 \]

### Step 4: Arrange A, I, and O
The letters A, I, and O can be arranged among themselves. The number of arrangements of 3 distinct letters is:
\[ 3! = 6 \]

### Step 5: Arrange Remaining Letters
After placing A, I, and O, we have 6 letters left: I, N, D, I, N, L. We need to arrange these 6 letters in the remaining positions (2 odd positions and 4 even positions).

The total number of arrangements of these letters, accounting for the repetition of I and N, is given by:
\[ \frac{6!}{2! \times 2!} = \frac{720}{4} = 180 \]

### Step 6: Calculate Total Arrangements
Now, we combine all the calculations:
- Ways to choose positions for A, I, O: 10
- Ways to arrange A, I, O: 6
- Ways to arrange the remaining letters: 180

Thus, the total number of permutations is:
\[ 10 \times 6 \times 180 = 10800 \]

### Final Answer
The number of permutations in which A, I, and O occur only at odd places is **10800**.

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