JEE Main 2023 · 8 April · morning · Single correct · +4 / −1
The number of arrangements of the letters of the word "INDEPENDENCE" in which all the vowels always occur together is :
- (a) 16800
- (b) 14800
- (c) 18000
- (d) 33600
Answer and solution
Answer (a) 16800
Solution
Maths sir · 4 steps
Glue the five vowels into one block, arrange the block with the consonants, then shuffle inside the block.
- Sort the letters. INDEPENDENCE has 12 letters. The vowels are I, E, E, E, E (5 letters, with E four times). The consonants are N, N, N, D, D, P, C (7 letters, with N three times and D twice).
- Arrange the units. Treat the 5 vowels as one block. Now you have 8 units: the block plus 7 consonants. Divide by 3! for the repeated N and 2! for the repeated D: 8!/(3! 2!) = 40320/12 = 3360.
- Inside the block. The block holds I and four identical E's, so it can be arranged in 5!/4! = 5 ways.
8!/(3! 2!) × 5!/4! = 3360 × 5 = 16800
- So. There are 16800 arrangements, which is option (a).
The idea to keep. For "always together", tie the group into one unit, arrange the units, then multiply by the arrangements inside the group. Divide out repeats at both stages.
Answer from the NTA final answer key. Solution by PrepWiser.