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

Instead of using the divide by repeats shortcut straight away, we can build the arrangement position by position, choosing spots for the repeated letters first. This is a slower but very concrete way to reach the same count, and it is a good check on the factorial method.

The word EXAMINATION has 11 letters. Splitting into vowels (E, A, A, I, I, O; 6 letters) and consonants (X, M, N, T, N; 5 letters), and noting that the 6 odd positions (1,3,5,7,9,11) exactly match the count of vowels, every vowel must go into an odd slot and every consonant into an even slot.

1. Placing the vowels among the 6 odd slots: first choose 2 of the 6 odd slots for the two A's, in $\binom{6}{2}=15$ ways. From the remaining 4 slots, choose 2 for the two I's, in $\binom{4}{2}=6$ ways. The last 2 slots take E and O, which can go in either order, in $2!=2$ ways. Multiplying these choices: $15 \times 6 \times 2 = 180$ ways to place the vowels.
2. Placing the consonants among the 5 even slots: first choose 2 of the 5 even slots for the two N's, in $\binom{5}{2}=10$ ways. The remaining 3 slots take X, M, and T in any order, in $3!=6$ ways. Multiplying: $10 \times 6 = 60$ ways to place the consonants.
3. Combining both halves: the vowel placement and the consonant placement happen in separate, non overlapping sets of slots, so every choice of vowel arrangement can be paired with every choice of consonant arrangement. Total arrangements $= 180 \times 60 = 10{,}800$.

This matches the factorial shortcut $\frac{6!}{2!\,2!} \times \frac{5!}{2!} = 180 \times 60 = 10{,}800$ exactly, confirming the count.

Let's summarize:

• 6 vowels must fill the 6 odd positions, in 180 distinct ways once repeats of A and I are accounted for.
• 5 consonants must fill the 5 even positions, in 60 distinct ways once the repeat of N is accounted for.
• The two placements are independent, so the totals multiply.

So the total number of valid arrangements is 10,800, option (C).