3 votes 3 votes How many ways can n books be placed on k distinguishable shelves if no two books are the same, and the positions of the books on the shelves matter? Combinatory kenneth-rosen discrete-mathematics combinatory + – aditi19 asked Apr 16, 2019 aditi19 1.6k views answer comment Share Follow See all 32 Comments See all 32 32 Comments reply Show 29 previous comments tusharp commented Apr 21, 2019 reply Follow Share k^n fails because: Take 3 shelves and 4 books 1. take the first book, you have 3 options. 2. For the second book, you don't have just 3 options. You have 3 ways to put it as the first book in any of the shelves OR to the right of the first book. 4 ways. So for every book options varies like k, k+1, k+2 and so on. 1 votes 1 votes Verma Ashish commented Apr 21, 2019 reply Follow Share Yes..correct👍👍 0 votes 0 votes mb14 commented Nov 2, 2019 i edited by mb14 Nov 2, 2019 reply Follow Share @tusharp I understood the case but how they derived the formula (k+n-1)!/(k-1)! for the product k,k+1,k+2..... why you are considering indistinguishable to distinguishable case when already said that no two books are same. Hence books are distinguishable. Anyone tell. 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes Source: Rosen Solution Manual smsubham answered Mar 1, 2020 smsubham comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes n books places at k places with no repeatation(no two books are the same) and the no two books are the same means permutations so answer is npk ma1999 answered Apr 20, 2019 ma1999 comment Share Follow See all 0 reply Please log in or register to add a comment.