If all vowels occupy odd places, how many words can be formed from the letters of the word HALLUCINATION?

Options

A. 129650
B. 1587600
C. 78500
D. none of these

Answer

To solve the problem of how many words can be formed from the letters of the word "HALLUCINATION" with the condition that all vowels occupy odd places, we will follow these steps:

### Step 1: Identify the letters in "HALLUCINATION"
The word "HALLUCINATION" consists of the following letters:
- Vowels: A, A, U, I, A, U, I, O (Total: 6 vowels)
- Consonants: H, L, L, C, N, T, N (Total: 7 consonants)

### Step 2: Determine the positions available for vowels
In a word of 13 letters, the odd positions are:
1st, 3rd, 5th, 7th, 9th, 11th, and 13th.
This gives us a total of 7 odd positions.

### Step 3: Choose positions for the vowels
Since we have 6 vowels and 7 odd positions, we need to select 6 out of these 7 positions. The number of ways to choose 6 positions from 7 is given by the combination formula:
\[ \binom{7}{6} = 7 \]

### Step 4: Arrange the vowels in the selected positions
The vowels we have are A, A, U, I, A, U, I, O. Since we have repetitions, we will use the formula for permutations of multiset:
\[ \text{Arrangements of vowels} = \frac{n!}{n_1! \times n_2! \times \cdots} \]
where \( n \) is the total number of vowels and \( n_1, n_2, \ldots \) are the counts of each unique vowel.

Here, we have:
- A appears 3 times
- U appears 2 times
- I appears 2 times
- O appears 1 time

Thus, the arrangements of the vowels:
\[ \text{Arrangements of vowels} = \frac{6!}{3! \times 2! \times 2! \times 1!} \]

### Step 5: Arrange the consonants
After placing the vowels, we have 7 consonants (H, L, L, C, N, T, N). The arrangements of the consonants will also consider repetitions:
\[ \text{Arrangements of consonants} = \frac{7!}{2! \times 2!} \]
where we divide by \( 2! \) for the two L's and \( 2! \) for the two N's.

### Step 6: Calculate the total arrangements
The total number of arrangements is given by multiplying the number of ways to choose positions for vowels, the arrangements of vowels, and the arrangements of consonants:
\[ \text{Total arrangements} = \binom{7}{6} \times \frac{6!}{3! \times 2! \times 2! \times 1!} \times \frac{7!}{2! \times 2!} \]

### Step 7: Simplify and calculate
Calculating each part:
1. \(\binom{7}{6} = 7\)
2. \(\frac{6!}{3! \times 2! \times 2! \times 1!} = \frac{720}{6 \times 2 \times 2 \times 1} = \frac{720}{24} = 30\)
3. \(\frac{7!}{2! \times 2!} = \frac{5040}{2 \times 2} = \frac{5040}{4} = 1260\)

Now, substituting back:
\[ \text{Total arrangements} = 7 \times 30 \times 1260 = 7 \times 37800 = 264600 \]

Thus, the total number of words that can be formed is **264600**.