3 votes 3 votes rajoramanoj asked Jan 11, 2018 rajoramanoj 318 views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments Ashwin Kulkarni commented Jan 11, 2018 reply Follow Share Case 1 : Out of 4 keys you can take 3 keys in one slot. which would lead to prob (1/8^3) and 1 remaining key will be in any of the remaining slot (7/8) Hence first case will be $^4C_3*\frac{7}{8^3}$ Case 2 : All four keys will be in one single slot . 0 votes 0 votes srivivek95 commented Jan 11, 2018 reply Follow Share shouldn't it be 84 instead of 83 ? @Ashwin 0 votes 0 votes rajoramanoj commented Jan 11, 2018 reply Follow Share same doubt .... 0 votes 0 votes Please log in or register to add a comment.