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How many disctict 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. $11!$
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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.