How many words can be formed with the letters of the word 'ANGLE' in which vowels occupy odd positions?

Answer

To solve the problem of how many words can be formed with the letters of the word "ANGLE" in which vowels occupy odd positions, we can follow these steps:

### Step 1: Identify the letters in the word "ANGLE"
The word "ANGLE" consists of 5 letters:
- Vowels: A, E
- Consonants: N, G, L

### Step 2: Determine the positions for vowels
In the word "ANGLE", the positions are:
1. 1st position (odd)
2. 2nd position (even)
3. 3rd position (odd)
4. 4th position (even)
5. 5th position (odd)

The odd positions are 1, 3, and 5. We need to place the vowels A and E in these positions.

### Step 3: Choose positions for the vowels
We need to choose 2 out of the 3 available odd positions for the vowels. This can be done using combinations:
\[ \text{Number of ways to choose 2 positions from 3} = \binom{3}{2} = 3 \]

### Step 4: Arrange the vowels in the chosen positions
Once we have chosen the positions for the vowels, we can arrange the 2 vowels (A and E) in those positions. The number of arrangements of 2 vowels is given by:
\[ 2! = 2 \]

### Step 5: Determine the positions for the consonants
After placing the vowels, we have 1 odd position left and 2 even positions (2nd and 4th) to fill with consonants (N, G, L).

### Step 6: Arrange the consonants
The remaining 3 positions (1 odd and 2 even) will be filled with the 3 consonants. The number of arrangements of 3 consonants is given by:
\[ 3! = 6 \]

### Step 7: Calculate the total arrangements
Now, we can find the total number of arrangements by multiplying the number of ways to choose the positions for the vowels, the arrangements of the vowels, and the arrangements of the consonants:
\[ \text{Total arrangements} = \binom{3}{2} \times 2! \times 3! = 3 \times 2 \times 6 = 36 \]

### Final Answer
The total number of words that can be formed with the letters of the word "ANGLE" in which vowels occupy odd positions is **36**.

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