0 votes 0 votes Anyone please give a solution to solve such question as it is difficult in one go. Algorithms hashing + – HeadShot asked Nov 30, 2018 • retagged Jun 11, 2022 by makhdoom ghaya HeadShot 888 views answer comment Share Follow See all 17 Comments See all 17 17 Comments reply arya_stark commented Nov 30, 2018 reply Follow Share Just check this https://gateoverflow.in/188101/hashing-test-series after seeing this i think your all doubt will be clear. 0 votes 0 votes OneZero commented Nov 30, 2018 reply Follow Share is it 20? 0 votes 0 votes arya_stark commented Nov 30, 2018 i edited by arya_stark Nov 30, 2018 reply Follow Share @OneZero i think its $72$ . leave $91$ and $77$ bcs they can come anytime. But $23$ Should come after $33$ and $44$ so first arrange $33$ and $44$ = $2!$ and after $ 23$, $ 64$ should come = $2! *1*1$ Now $91$ can come any time --> $6 Ways$ Same for 77 --> $6 Ways$ So , $Total$ $Possible$ $Ways$ $= 6*6*2= 72$ Correct me if i wrong!!! 0 votes 0 votes HeadShot commented Nov 30, 2018 reply Follow Share @arya_stark https://gateoverflow.in/188101/hashing-test-series?show=188222#a188222 in this case 2. why 91 can not come before 12 ? there is no dependency between 12 and 91 so ? p.s : For this que ans is 60. 0 votes 0 votes OneZero commented Nov 30, 2018 i edited by OneZero Nov 30, 2018 reply Follow Share @arya_stark you cannot place 64 before 23 because 64 will take position 5 also after arranging 33 44 23 and 64 you are left with 5 spaces between then i.e.. __ 33 __ 44 __ 23 __ 64 __ 91 and 77 can go in any of there places hence 5*4 final answer according to me should me 5*4*2 = 40. my first answer 20 is wrong. sorry for that. 0 votes 0 votes HeadShot commented Nov 30, 2018 reply Follow Share @OneZero answer this : https://gateoverflow.in/272384/hashing-counting?show=272423#c272423 0 votes 0 votes OneZero commented Nov 30, 2018 reply Follow Share @HeadShot you linked to this page itself. 0 votes 0 votes HeadShot commented Nov 30, 2018 reply Follow Share @OneZero yes. Please check that comment . 0 votes 0 votes arya_stark commented Nov 30, 2018 reply Follow Share Pardon me @HeadShot @OneZero !! I did big mistake. Lets take one by one 1) 33 comes before 23 and 44 comes before 64 So, sequence (33,44) comes before (23,64) Here 33 and 44 can come like :: (33,44) or (44,33) means => 2! But 23 and 64 should come in fixed 23 then 64. 2) Lets Consider 91 : it can come any place $_ ,33,_,44,_ ,23,_64_$ ==> $5$ $Ways$ 3) Now for 77 : $_,91,_,33,_,44,_,23,_,64_$ ==> $6$ $Ways$ $Total $Possible$ $Way$ = $2*5*6 = 60$ 1 votes 1 votes Subarna Das commented Nov 30, 2018 reply Follow Share @HeadShot, Please add question source in title 1 votes 1 votes OneZero commented Nov 30, 2018 reply Follow Share sorry for all that mess. got the answer! so we know that 23 and 64 should come after 33 and 44 also 23 should come before 64 or else 64 will take position 5. so now we have __ 33 __ 44 __ 23 __ 64 __ or __ 44 __ 33 __ 23 __ 64 __ therefore we have 2 options. now place either 91 or 77 into there blanks. (anywhere in these blanks). suppose we place 91 into one these blanks (it can be placed in 5 ways). the outcome would be __ 33 __ 91 __ 44 __ 23 __ 64 __ we can see that we have 6 blanks spaces for 77 to be placed. hance answer is 2*5*6 = 60. 0 votes 0 votes arya_stark commented Nov 30, 2018 reply Follow Share @OneZero Yoo Bro 😂🤟 0 votes 0 votes OneZero commented Nov 30, 2018 reply Follow Share LOL. finally. 0 votes 0 votes HeadShot commented Nov 30, 2018 reply Follow Share @arya_stark @OneZero Guys please answer this comment too 😅 https://gateoverflow.in/272384/hashing-counting?show=272423#c272423 0 votes 0 votes OneZero commented Nov 30, 2018 reply Follow Share @HeadShot He solved it in lengthy way. so we need to have {2,13,24} before 12,62 and 82 also we need to have 77 placed before 82. now {2,13,24} can be arranged in 3! ways. for representing further steps i will take the help of the below representation (this is one of the 3! possible representation of {2,13,24}) __ 2 __ 13 __ 24 __ 12 __ 62 __ 82 __ we know that 77 should come before 82 hence the possible places are __ 2 __ 13 __ 24 __ 12 __ 62 __ 82 i.e.. 6 places the following is one of the 6 representations __ 2 __ 13 __ 77 __ 24 __ 12 __ 62 __ 82 __ now we have 8 possible places to place 91. hence answer is 3!*6*8 = 288 0 votes 0 votes pradeepchaudhary commented Dec 2, 2018 reply Follow Share Where is 12 ?? 0 votes 0 votes arya_stark commented Dec 2, 2018 reply Follow Share @pradeepchaudhary it is explanation of this question .... https://gateoverflow.in/188101/hashing-test-series 0 votes 0 votes Please log in or register to add a comment.