SOLUTION: 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

Algebra ->  Permutations -> SOLUTION: 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       Log On


   



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(52781) About Me  (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.

--------------------

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.


/\/\/\/\/\/\/\/

Do not forget to post your  " THANKS "  to me for my teaching  (!)