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| Question 1173343:  From the numbers 1,2,3,4,5,6,7,8, and 9, four different numbers are selected to form a four-digit number.
 a.How much four-digit numbers can be formed?
 b.How many four-digit less than 2,000 can be formed?
 c.How many four-digit even numbers can be formed?
 Answer by ikleyn(52879)
      (Show Source): 
You can put this solution on YOUR website! . 
 
(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".
 
 
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