SOLUTION: In how many ways 7 different books can be arranged on a shelf? In how man many ways will a specified book be (I) at the centre? (II) at either end?

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Question 1114136: In how many ways 7 different books can be arranged on a shelf? In how man many ways will a specified book be (I) at the centre? (II)
at either end?

Answer by ikleyn(52777)   (Show Source): You can put this solution on YOUR website!
.
(I)  It is VERY STANDARD introductory problem on PERMUTATIONS for beginners.


     The answer is  7! = 1*2*3*4*5*6*7.


     The standard solution (=mantra) is this:


         You can put any of the 7 books in the 1-st position in 7 ways.

         You can put any of the remaining 6 books in the 2-nd position in 6 ways.
         
         You can put any of the remaining 5 books in the 3-rd position in 5 ways.


         . . . and so on . . . 


         You can put any of the remaining 2 books in the 6-th position in 2 ways.


         Finally, you can put the last remaining 1 book in the 7-th position in 1 way only (there is nothing to select from at this step . . . ).

         
         Since these selections/options are independent, you have  7! = 7*6*5*4*3*2*1  ways, in all.

Solved.

-----------------
On Permutations,  see the lessons
    - Introduction to Permutations
    - PROOF of the formula on the number of Permutations
    - Problems on Permutations

    - OVERVIEW of lessons on Permutations and Combinations
in this site.

Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic  "Combinatorics: Combinations and permutations".


Save the link to this textbook together with its description

Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson

into your archive and use when it is needed.


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