1 votes 1 votes Given symbols:- $GGGGGAAATTTECCS$ No of ways such that exactly 4 Gs out of 5 Gs are together. I am getting$ \frac{11! × 11}{3!×3!×2!} $. Some one verify it? Combinatory combinatory engineering-mathematics + – Jason asked Apr 12, 2018 Jason 692 views answer comment Share Follow See all 18 Comments See all 18 18 Comments reply Show 15 previous comments srestha commented Apr 12, 2018 reply Follow Share then why not ans is $\frac{12!}{3!3!2!}$ that means there will be always atleast a place between GGGG_G and that place should be fill up with other element 0 votes 0 votes Tesla! commented Apr 12, 2018 reply Follow Share Yes and that gap can be filled with single or multiple elements And 12! Can have atleast one permutation where 5g will be present together 0 votes 0 votes Utkarsh Joshi commented Aug 2, 2018 reply Follow Share What about treating 4G's as a single unit and considering that as just one G?? it would give 12! / ( 2! * 3! * 3! *2!) permutations. 0 votes 0 votes Please log in or register to add a comment.