.
Any one of 13 volumes she can place to the first position;
any one of remaining 12 volumes she can place to the second position;
any one of remaining 11 volumes she can place to the third position,
. . . . . . . . and so on . . . . .
till the last, 13-th volume;
which gives her 13! = 13*12*11*10*9*8*7*6*5*4*3*2*1 options, in all.
+-----------------------------------------------------------------+
| |
| Any set of n different objects has n! different permutations. |
| |
+-----------------------------------------------------------------+
Congratulations !
For the first time in your life, I presented you this famous basic problem on PERMUTATIONS (!)
Happy learning (!)
------------------
On Permutations, see introductory 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.