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+7 votes
163 views

How many distinct words can be formed by permuting the letters of the word ABRACADABRA?

  1. $\frac{11!}{5! \: 2! \: 2!}$
  2. $\frac{11!}{5! \: 4! }$
  3. $11! \: 5! \: 2! \: 2!\:$
  4. $11! \: 5! \: 4!$
  5. $11! $
asked in Combinatory by Veteran (99.2k points)
edited ago by | 163 views

3 Answers

+10 votes
Best answer
$\text{ABRACADABRA}$

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

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.
answered by Veteran (21.6k points)
edited ago by
+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.
answered by Loyal (4.6k points)
+1 vote

The correct answer is (A) 11! / (5!⨉2!⨉2!)

answered by Veteran (13.7k points)


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