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5 different items can be ordered in 5! = 5*4*3*2*1 = 120 different ways.
Any of 5 items can be placed in 1st position.
Any of remaining 4 items can be placed in 2nd position.
Any of remaining 3 items can be placed in 3rd position.
Any of remaining 2 items can be placed in 4th position.
The last remaining item can be placed in the last, 5th position.
Counting this way, you get the formula and the value.
Solved and explained.
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This problem is on PERMUTATIONS.
On Permutations, see introductory lessons
- Introduction to Permutations
- PROOF of the formula on the number of Permutations
- Simple and simplest problems on permutations
- Special type permutations problems
in this site.
Learn the subject from there ( ! )
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.