Question

All the letters of the word 'ABRACADABRA' are arranged in different possible ways. Then the number of such arrangements in which the vowels are together is

• A. 1200
• B. 1240
• C. 1220
• D. 1260

Hint:

Whenever a string counting problem says certain elements must be "together", always treat them as a single rigid block element! This collapses an 11-letter problem down to a much simpler 7-element permutation layout.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to find the total number of unique permutations of the letters in the word 'ABRACADABRA' such that all vowel letters are clustered tightly together as a single group.

Step 2: Detailed Explanation:
Let's first analyze and count the individual letter repetitions in the 11-letter word 'ABRACADABRA':

• Vowels: A, A, A, A, A $\rightarrow$ Total of 5 occurrences of the letter 'A'.

• Consonants: B, R, C, D, B, R $\rightarrow$ Total of 6 letters (where 'B' repeats 2 times, 'R' repeats 2 times, 'C' appears 1 time, and 'D' appears 1 time).
Since the condition requires all vowels to remain together, we tie the 5 'A's together into a single composite entity or "super-letter": [AAAAA]. Now, we count the total number of independent entities left to arrange: $$ \text{Total Entities} = 6 \text{ consonants} + 1 \text{ vowel block} = 7 \text{ entities} $$ Let's calculate the permutations of these 7 entities, accounting for the repetitions among the consonants ('B' repeats 2 times, 'R' repeats 2 times): $$ \text{Arrangements} = \frac{7!}{2! \times 2!} $$ $$ \text{Arrangements} = \frac{5040}{2 \times 2} = \frac{5040}{4} = 1260 $$ Inside the vowel block [AAAAA], all 5 letters are identical 'A's, meaning there is only exactly 1 unique way to arrange them internally ($\frac{5!}{5!} = 1$). Thus, the total number of valid structural arrangements is: $$ 1260 \times 1 = 1260 $$

Step 3: Final Answer:
The number of such arrangements where vowels remain together is 1260, corresponding to option (D).