JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Permutations and Combinations

The word INDEPENDENCE has 12 letters: vowels I, E, E, E, E (1 I and 4 E's) and consonants N, N, N, D, D, P, C (3 N's, 2 D's, 1 P, 1 C). Treating all vowels as a single unit, we have 8 units to arrange: (vowels), N, N, N, D, D, P, C. These can be arranged in $\frac{8!}{3!\cdot 2!}$ ways (dividing by $3!$ for the 3 N's and $2!$ for the 2 D's). The vowels within their unit can be arranged in $\frac{5!}{4!}=5$ ways (dividing by $4!$ for the 4 E's). Total arrangements = $\frac{8!}{3!\cdot 2!}\times 5=\frac{40320}{6\cdot 2}\times 5=\frac{40320}{12}\times 5=3360\times 5=16800$.