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. $11!$
edited ago | 163 views

$\text{ABRACADABRA}$

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

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 ago
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.
+1 vote