Question:

Step 1: Identify the letters in 'CHITRANJEEVI': C, H, I, T, R, A, N, J, E, E, V, I. Total = 12 letters. Vowels: A, E, E, I, I (5 vowels with E repeated twice and I repeated twice). Consonants: C, H, T, R, N, J, V (7 consonants, all distinct). Step 2: Option (a) — All vowels separated: Arrange 7 consonants in $7!$ ways. This creates 8 gaps. Choose 5 gaps from 8: ${}^8C_5$ ways. Arrange 5 vowels (A, E, E, I, I) in chosen gaps: $\frac{5!}{2!2!}$ ways. Total = $7! \times {}^8C_5 \times \frac{5!}{2!2!}$. Option (a) is correct. Step 3: Option (b) — Vowels in alphabetical order (A, E, E, I, I): Total arrangements of 12 letters with repetitions = $\frac{12!}{2!2!}$. The 5 vowels can be arranged in $\frac{5!}{2!2!}$ ways, of which exactly 1 is in alphabetical order. So number of words = $\frac{12!}{2!2!} \times \frac{1}{\frac{5!}{2!2!}} = \frac{12!}{2!2!} \times \frac{2!2!}{5!} = \frac{12!}{5!}$. Option (b) is correct. Step 4: Option (c) — Words containing 'CHITRA': Treat 'CHITRA' as a single block. Remaining letters: N, J, E, E, V, I, I (7 letters with E, E and I, I repeated). Total arrangements = $\frac{7!}{2!2!}$... Wait, the option states $\frac{7!}{2!}$. Remaining letters after removing C,H,I,T,R,A from CHITRANJEEVI: N, J, E, E, V, I (6 letters + 1 block = 7 units). Letters: N, J, E, E, V, I — here E repeats twice. So arrangements = $\frac{7!}{2!}$. Option (c) is correct. Step 5: Option (d) — Words containing 'IITJEE': Treat 'IITJEE' as a single block. Remaining letters from CHITRANJEEVI after removing I,I,T,J,E,E: C, H, R, A, N, V (6 letters, all distinct). Total units = 6 + 1 = 7, all distinct. Arrangements = $7!$. Option (d) is correct.