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 find the number of permutations of the letters in the word "INDIANOIL" such that the letters A, I, and O occur only at odd places, we can follow these steps:

### Step 1: Identify the Total Letters and Their Frequencies
The word "INDIANOIL" consists of 9 letters: I, N, D, I, A, N, O, I, L. The frequency of each letter is:
- I: 3
- N: 2
- D: 1
- A: 1
- O: 1
- L: 1

### Step 2: Identify the Odd Places
In a 9-letter arrangement, the odd positions are 1, 3, 5, 7, and 9. This gives us a total of 5 odd positions.

### Step 3: Place A, I, and O in Odd Positions
We need to place the letters A, I, and O in the odd positions. Since we have 5 odd positions and we need to choose 3 of them for A, I, and O, we can choose any 3 odd positions from the 5 available.

The number of ways to choose 3 positions from 5 is given by:
\[ \text{Ways to choose 3 positions} = \binom{5}{3} = 10 \]

### Step 4: Arrange A, I, and O in the Chosen Positions
The letters A, I, and O can be arranged in the chosen positions. Since I repeats 3 times, the arrangements of A, I, and O in the chosen positions is given by:
\[ \text{Arrangements of A, I, O} = \frac{3!}{1! \cdot 2!} = 3 \]

### Step 5: Fill the Remaining Positions
After placing A, I, and O, we have 6 positions left (2 odd and 4 even). The remaining letters are I, N, N, D, L (since we have already used one A, one I, and one O).

The number of ways to arrange these remaining letters is given by:
\[ \text{Arrangements of remaining letters} = \frac{6!}{2!} = \frac{720}{2} = 360 \]

### Step 6: Calculate the Total Permutations
Now, we can calculate the total permutations by multiplying the number of ways to choose positions, the arrangements of A, I, O, and the arrangements of the remaining letters:
\[ \text{Total permutations} = \text{Ways to choose positions} \times \text{Arrangements of A, I, O} \times \text{Arrangements of remaining letters} \]
\[ = 10 \times 3 \times 360 = 10800 \]

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

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