Question

In how many ways can the letters of the word POTICA be arranged such that the vowels occupy odd positions?

• A. 24
• B. 30
• C. 36
• D. 42

Hint:

When positions are constrained (odd/even), first lock the allowed slots for the restricted group, permute within groups, then multiply.

Answer 1

The Correct Option is C

Letters: P, O, T, I, C, A (all distinct). Vowels \(= \{O,I,A\}\) (3), consonants \(=\{P,T,C\}\) (3). Positions are \(1,2,3,4,5,6\); odd positions are \(1,3,5\). Step 1: Place the vowels in the odd slots.
All three odd slots must be filled by the three vowels. They can be permuted in \(3!\) ways. Step 2: Place the consonants in the even slots.
Even positions \(2,4,6\) are filled by P, T, C in \(3!\) ways. Step 3: Multiply independent choices.
Total arrangements \[ 3!\times 3!=6\times 6=36. \] Sanity check (alternate view).
Total permutations of 6 distinct letters is \(6!=720\). Probability a random permutation has vowels in the three odd slots equals \[ \frac{\binom30\,\binom33}{\binom63} \cdot \frac{3!\,3!}{6!}=\frac{1\cdot1}{20}\cdot \frac{36}{720}=\frac{36}{720}, \] which again gives \(36\) favorable arrangements. \[ \boxed{36} \]