Question

How many arrangements can be formed out of the letters of the word EXAMINATION so that vowels always occupy odd places?

• A. 72,000
• B. 86,400
• C. 10,800
• D. 64,000

Hint:

Count the odd positions (6 of them) and notice it equals the number of vowels, so vowels and consonants split cleanly into two separate groups to arrange.

Answer 1

The Correct Option is C

Step 1: List the letters and split vowels from consonants.
The word EXAMINATION has 11 letters: E, X, A, M, I, N, A, T, I, O, N.
Vowels: E, A, A, I, I, O, a total of 6 vowels, with A repeated twice and I repeated twice.
Consonants: X, M, N, T, N, a total of 5 consonants, with N repeated twice.

Step 2: Count the odd positions.
In an 11 letter arrangement, the positions are numbered 1 to 11. The odd positions are 1, 3, 5, 7, 9, 11, exactly 6 positions, the same as the number of vowels. So every vowel must sit in an odd position, and every consonant must sit in one of the remaining 5 even positions (2, 4, 6, 8, 10).

Step 3: Arrange the vowels in the 6 odd positions.
The 6 vowels are E, A, A, I, I, O, with A appearing twice and I appearing twice. The number of distinct arrangements of these 6 letters is
\[ \frac{6!}{2! \times 2!} = \frac{720}{4} = 180 \]

Step 4: Arrange the consonants in the 5 even positions.
The 5 consonants are X, M, N, T, N, with N appearing twice. The number of distinct arrangements is
\[ \frac{5!}{2!} = \frac{120}{2} = 60 \]

Step 5: Combine the two independent choices.
Since the vowel arrangement and the consonant arrangement happen independently of each other, the total number of arrangements is the product:
\[ 180 \times 60 = 10{,}800 \]

Final Answer:
\[ \boxed{10{,}800} \]