0 votes 0 votes consider 5 numbered balls and 5 numbered boxes , how many ways are possible such that exactly 2 balls is not on the ith spot ? [Ball i is in the ith box ] hitendra singh asked Oct 27, 2018 hitendra singh 841 views answer comment Share Follow See all 12 Comments See all 12 12 Comments reply minal commented Oct 27, 2018 i edited by minal Oct 27, 2018 reply Follow Share 2!*5c2(1-1+1/2)=10 ? 0 votes 0 votes hitendra singh commented Oct 27, 2018 reply Follow Share i didn't get it can you explain a bit? 0 votes 0 votes Balaji Jegan commented Oct 27, 2018 i edited by Balaji Jegan Oct 27, 2018 reply Follow Share Can anyone tell what is wrong in this? It is exceeding total arrangements of 5! = 120 ways. 1st let us select 2 balls => 5C2 = 10 The 1st ball has 4 positions. If the 1st ball occupies the position which must not be occupied by 2nd ball, then 2nd ball has 4 positions otherwise if the 1st ball occupies any other position, then 2nd ball has only 3 positions. The remaining 3 balls can be arranged in 3P3 ways = 3! ways = 6 ways Thus total ways = 10 * ((1*4)+(3*3)) * 6 = 10*(4+9)*6 = 780 ways 0 votes 0 votes hitendra singh commented Oct 27, 2018 reply Follow Share if the question would have been like this what would be the answer ? ways such that ball i is in the ith box? 0 votes 0 votes minal commented Oct 27, 2018 reply Follow Share only 1 way https://en.wikipedia.org/wiki/Derangement 0 votes 0 votes hitendra singh commented Oct 27, 2018 reply Follow Share @ Balaji Jegan if you arrange rest three balls in 3! ways then these contains permutation which will have some balls on the correct position which will increase the count on number of balls on the correct place . I hope u understand What I wanted to say here? 1 votes 1 votes Utkarsh Joshi commented Oct 27, 2018 reply Follow Share Question is incomplete. Can a box remain empty? 1 votes 1 votes hitendra singh commented Oct 27, 2018 reply Follow Share @ Utkarsh Joshi no box remains empty . 0 votes 0 votes Utkarsh Joshi commented Oct 27, 2018 reply Follow Share Then it would be 5c2=10. 1 votes 1 votes Utkarsh Joshi commented Oct 27, 2018 reply Follow Share 10 is correct,isn't it? 0 votes 0 votes Vikas Verma commented Oct 27, 2018 reply Follow Share Yeah even I think it should be 10. Put all balls in their respective boxes. Select any 2 boxes and swap their balls. 0 votes 0 votes hitendra singh commented Oct 27, 2018 reply Follow Share yeah i think so thanks! 0 votes 0 votes Please log in or register to add a comment.