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(a) 5 different letters admit 5! = 5*4*3*2*1 = 120 PERMUTATIONS.
(b) 5 letters, with one letter "P" of the multiplicity 2 admit = = 60 different arrangements.
Solved.
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On permutations, see the lessons
- Introduction to Permutations
- PROOF of the formula on the number of Permutations
- Simple and simplest problems on permutations
- Arranging elements of sets containing indistinguishable elements
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.