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.
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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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