.
"Less than 2,000" means that the first digit is " 1 " (thousand's digit).
So, first digit is fixed; it is the digit " 1 ".
In the rest 3 positions, any of 8 digits from 2 to 9 can be.
It gives 8*8*8 = = 512 opportunities and 512 possible 4-digit numbers,
under given condition.
Solved.
This solution is produced assuming that repeating is allowed.
If repeating is not allowed, then the answer is 8*7*6 = 336.
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If you want to see many similar solved problems on PERMUTATIONS, see my 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.