How many words can be formed out of the letters of the word education such that vowels occupy the odd positions?

Answer

To solve the problem of how many words can be formed out of the letters of the word "education" such that vowels occupy the odd positions, we can follow these steps:

### Step 1: Identify the letters in the word "education"
The word "education" consists of 9 letters: E, D, U, C, A, T, I, O, N.

### Step 2: Identify the vowels and consonants
In the word "education", the vowels are:
- E
- U
- A
- I
- O

This gives us a total of **5 vowels**.

The consonants in the word are:
- D
- C
- T
- N

This gives us a total of **4 consonants**.

### Step 3: Determine the positions for the vowels
Since we need the vowels to occupy the odd positions in a 9-letter word, the odd positions are:
1, 3, 5, 7, 9

This means there are **5 odd positions** available for the 5 vowels.

### Step 4: Arrange the vowels in the odd positions
The number of ways to arrange the 5 vowels in the 5 odd positions is given by the factorial of the number of vowels:
\[ 5! = 120 \]

### Step 5: Arrange the consonants in the remaining positions
The remaining positions (even positions) for the consonants are:
2, 4, 6, 8

This gives us **4 positions** for the 4 consonants.

The number of ways to arrange the 4 consonants in these 4 positions is given by the factorial of the number of consonants:
\[ 4! = 24 \]

### Step 6: Calculate the total arrangements
To find the total number of arrangements of the letters in the word "education" with the vowels in odd positions, we multiply the number of arrangements of vowels by the number of arrangements of consonants:
\[ \text{Total arrangements} = 5! \times 4! = 120 \times 24 = 2880 \]

### Final Answer
Thus, the total number of words that can be formed is **2880**.

---