15 votes 15 votes How many distinct words can be formed by permuting the letters of the word $\text{ABRACADABRA}?$ $\frac{11!}{5! \: 2! \: 2!}$ $\frac{11!}{5! \: 4! }$ $11! \: 5! \: 2! \: 2!\:$ $11! \: 5! \: 4!$ $11! $ Combinatory tifr2017 combinatory counting easy + – go_editor asked Dec 21, 2016 • retagged Nov 21, 2022 by Lakshman Bhaiya go_editor 2.2k views answer comment Share Follow See 1 comment See all 1 1 comment reply Warrior commented Aug 30, 2017 i edited by Satbir Jun 27, 2019 reply Follow Share The correct answer is (A) 11! / (5!⨉2!⨉2!) 0 votes 0 votes Please log in or register to add a comment.
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. focus _GATE answered Dec 21, 2016 • edited May 20, 2018 by kenzou focus _GATE comment Share Follow See all 0 reply Please log in or register to add a comment.
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. junk_mayavi answered Dec 21, 2016 junk_mayavi comment Share Follow See all 0 reply Please log in or register to add a comment.