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: Count the letters.
The word ABRACADABRA has 11 letters. The vowels are five A's. The consonants are B, R, C, D, B, R, that is six letters with B twice and R twice.

Step 2: Tie the vowels into one block.
Since all vowels must stay together, treat the five A's as a single bundle. Now we arrange 6 consonants plus 1 vowel bundle, which is 7 items.

Step 3: Arrange the 7 items.
The consonants repeat (B twice, R twice), so \[ \frac{7!}{2!\,2!}=\frac{5040}{4}=1260 \]

Step 4: Arrange inside the block.
The five A's are identical, so there is only 1 way inside the bundle. The total stays $1260$. \[ \boxed{1260} \]