Rosen 7e Exercise-6.5 question 45.b page 433

Question

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?

Comments

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.

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Yes..correct👍👍

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@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. 

Answer 1

Source: Rosen Solution Manual

 

Answer 2

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