.
(a) 9 options for the most-left digit;
8 remaining options for the next digit;
7 remaining options for the next digit;
6 options for the last digit.
In all, there are 9*8*7*6 = 3024 such numbers. ANSWER
(b) " Less than 2000 " means that the first digit is 1.
Then there are 8 remaining options for the next digit;
7 remaining options for the next digit;
6 remaining options for the last digit.
In all, there are 8*7*6 = 336 such numbers. ANSWER
(c) The last digit ("ones" digit) is any of four even digits 2, 4, 6 8, giving 4 options
The "tens" digit is any of 8 remaining digits, giving 8 options;
The "hundreds" digit is any of 7 remaining digits, giving 7 options;
The "thousands" digit is any of 6 remaining digits, giving 6 options.
In all, the number of options is 4*8*7*6 = 1344, giving 1344 possible numbers.
Solved - all questions are answered.
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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.
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