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The total number of all possible permutations of 5 digits is T = 5! = 5*4*3*2*1 = 120.
The total number of those favorable permutations that start with three odd digits 1, 3 and 7, is F = (3*2*1)*2*1.
The probability under the question is the ratio
P = = = = = = = 0.1 = 10%. ANSWER
Solved.
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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
- Problems on Permutations with restrictions
- 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.