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15 votes
15 votes

How many distinct words can be formed by permuting the letters of the word $\text{ABRACADABRA}?$

  1. $\frac{11!}{5! \: 2! \: 2!}$
  2. $\frac{11!}{5! \: 4! }$
  3. $11! \: 5! \: 2! \: 2!\:$
  4. $11! \: 5! \: 4!$
  5. $11! $
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2 Answers

Best answer
14 votes
14 votes

$\text{ABRACADABRA}$

$A\rightarrow 5\\ B\rightarrow 2\\ R\rightarrow 2$

Total Permutation of words $={11!}$

Now,we have to remove word from total permutation of words which have repetition of letter,

$=\dfrac{11!}{5!2!2!}$

Hence option (A) is correct.

edited by
6 votes
6 votes
The word 'ABRACADABRA' have 11 letters. Therefore total permutations is 11!

However they are not unique 11 letters and have duplicates repeated as follows.

A - 5 times

B- 2 times

R- 2 times

C and D are unique.

Therefore answer will be option A.
Answer:

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