1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Need Help w/ Beginner Permutations Question

  1. Aug 14, 2014 #1
    1. The problem statement, all variables and given/known data
    "A 3-digit number is made up using the digits 0, 1, 2, 3, 4, 5, 6 and 7 at most once each. The number cannot start with 0. How many such numbers can be formed if:
    a. there are no other restrictions
    b. the number ends in a 5
    c. the number ends in a 0
    d. the number is divisible by 5 ?"

    2. The attempt at a solution

    a:
    Three digits: _ _ _
    First number can't be 0, 7 other possibilities: 7 _ _
    Second number can be 0, 1 number has been used (of the 8), 7 more possibilities: 7 x 7 _
    Third number can be any remaining 6: 7 x 7 x 6 = 294 Correct Answer is 180

    b:
    Three digits: _ _ _
    First number can't be 0, 5 is used, 6 other possibilities: 6 _ _
    Second number can be 0, 2 numbers used, 6 other possibilities: 6 x 6 _
    Third Number is 5: 6 x 6 x 1 = 36 Correct Answer is 25

    c:
    Three digits: _ _ _
    First number can't be 0, 0 is used, 7 other possibilities: 7 _ _
    Second number can't be 0, 2 numbers used, 6 other possibilities: 7 x 6 _
    Third Number is 0: 7 x 6 x 1 = 42 Correct Answer is 30

    d:
    Three digits: _ _ _
    First number can't be 0, 7 other possibilities: 7 _ _
    Second number can be 0, 7 remaining possibilities: 7 x 7 _
    Third number is either 0 or 5 (divisible by 5): 7 x 7 x 2 = 98 Correct Answer is 55

    I am sorry for this post being so large, but that is all my work. I am probably missing some simple step that is messing up all my answers, however, that is how my book has taught me to look at the problems. Any help with what I'm doing incorrectly would be greatly appreciated!
     
    Last edited: Aug 14, 2014
  2. jcsd
  3. Aug 14, 2014 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    It looks to me like the 'correct answers' are incorrect! I agree with your answers.
     
  4. Aug 14, 2014 #3

    verty

    User Avatar
    Homework Helper

    I get a different answer for d), I get 84 (mentally so this may be wrong).
     
  5. Aug 14, 2014 #4
    I agree with the others, the "correct" answers seem wrong.
     
  6. Aug 14, 2014 #5

    Nathanael

    User Avatar
    Homework Helper

    Wouldn't answer D be the sum of B and C?


    P.S.
    Are you sure the digits go up to 7?

    If the digits only go up to 6, then the given answers are correct, otherwise they're wrong.
     
    Last edited: Aug 14, 2014
  7. Aug 14, 2014 #6

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    I get 78 for case (d).
     
  8. Aug 14, 2014 #7

    verty

    User Avatar
    Homework Helper

    This is the correct answer.
     
  9. Aug 14, 2014 #8
    Perhaps instead of "correct answer" I should have put "book answer." I do not agree with the book's answers, and now that I have others who also disagree with them I think I'll e-mail my math teacher for clarification over the matter. Thanks!
     
  10. Aug 14, 2014 #9
    Also thanks everyone for your clarifications on case d, as I didn't think of using the sum principle because I wasn't paying enough attention and just kept going with the product principle.
     
  11. Aug 15, 2014 #10

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Did you notice Nathanael's (edited) post #5? All the book answers correspond to the numbers going up to 6, not 7.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Need Help w/ Beginner Permutations Question
  1. Permutations help (Replies: 3)

  2. Permutations question (Replies: 7)

  3. Permutation question (Replies: 3)

Loading...