TIFR CSE 2017 | Part A | Question: 6

Question

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

• A. $\frac{11!}{5! \: 2! \: 2!}$
• B. $\frac{11!}{5! \: 4! }$
• C. $11! \: 5! \: 2! \: 2!\:$
• D. $11! \: 5! \: 4!$
• E. $11! $

Comments

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

Answer 1

$\text{ABRACADABRA}$

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

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.

Answer 2

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.