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Fundamental counting principle

  1. Jul 3, 2016 #1
    1. The problem statement, all variables and given/known data
    How many 3 digit numbers can be formed from 0 to 3 without repition



    2. Relevant equations


    3. The attempt at a solution

    What I did first is

    H = number of choices for the hundreds place
    T = number of choices for the tens place
    U = number of choices for the units digit

    H = 3 possible choices (excluding zero)
    T = 3 possible choices ( including zero but excluding whatever is in the hundreds place)
    U = 2 possibe choises (including zero but excluding whatever are in the hundreds place tens place)

    Therefore 3x3x2 = 18 numbers can be formed

    But I noticed something. When do the selection in reverese that is selecting the choices for units digit first then the tens place then hundreds place I get a different answer

    U = number of choices for the units digit
    T = number of choices for the tens place
    H = number of choices for the hundreds place

    U = 4
    T= 3
    H = 1

    Therefore 4x3x1 = 12 numbers

    Can you guys explain why this happens? Thanks!
     
  2. jcsd
  3. Jul 3, 2016 #2

    SammyS

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    It's more difficult to do it if you don't do the hundred's place first.

    What to use for H depends upon whether or not a zero has been chosen for the unit's place or ten's place. You could find the total number of 3 digit numbers if you allow a zero in the hundred's place. Then subtract the number of three digit numbers in that list which have a zero in the offending position.
     
  4. Jul 3, 2016 #3
    If you use 0 for U or T you do have
    U = 4
    T = 3
    H = 1
    4x3x1 = 12 possibilities

    But, you also may not use 0 at all. Then
    U = 3
    T = 2
    H = 1
    3x2x1 = 6 possibilities

    The total is 12+6 = 18 possibilities
     
  5. Jul 3, 2016 #4
    Hello thank you for this reply.

    What if we are aksed
    How many 3 digit even numbers can be formed from 0, 1, 2, 3, 4, 5 and 6 with no repetition?
    My attempt:
    For U=0, T=6 choices, H = 5 choices
    30 numbers ending with zero not including two digit numbers starting with zero
    For U=2, T=6 choices, H = 4choices
    24 Numbers ending with two not including two digit numbers starting with zero
    For U=4, T=6 choices, H = 4 choices
    24 numbers ending with four not including two digit numbers starting with zero
    For U=6, T=6 choices, H = 4 choices
    24 numbers ending with 6 not including two digit numbers starting with zero
    ∴ 30+72=102 three-digit even numbers.

    The correct answer is 105 if we do the selection in reverse that is Tens place first the the hundreds

    How do we reconcile this to what you posted above? Thanks.
     
  6. Jul 3, 2016 #5

    SammyS

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    A couple comments are in order here.

    I really liked Irene Kaminkowa's method. And by the way: Welcome to PF, Irene.

    To avoid confusion we should make a distinction as to whether we are referring to a particular digit being used in some position, or the referring to the number of choices for that position. Thus I suggest using Upper Case for the latter as has been in the OP, and using lower case to refer to the value being considered for a particular position. E.g.: If u = 0, then there are 6 choices for t, and thus T = 6.
     
  7. Jul 4, 2016 #6
    Let's generalize:
    Given digits 0 to n, how many 3-digit numbers can we get without digit repetitions?

    From the search above we see that it's better to start from H. Anyway, the answer is the same. So
    H = n
    T = n
    U = n-1
    The number of possibilities
    NP(n) = n2 (n-1) - general formula

    Test for n = 3
    NP(3) = 9*2 = 18 - works!

    For n = 6
    NP(6) = 36*5 = 210 - all numbers, both even and odd
    So
    NP(6) / 2 = 105 - even 3-digit numbers containing digits from 0 to 6
     
    Last edited: Jul 4, 2016
  8. Jul 4, 2016 #7

    SammyS

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    (6)(6)(5) = 180, not 210 .

    Beyond that, why can we simply divide by 2 to get the number of even numbers. There are 7 possible digits here, 4 of which are even.
     
  9. Jul 4, 2016 #8
    Thank you, SammyS
    Let me correct my calculations.

    So, we have
    NP(6) = 36*5 = 180 numbers
    These numbers end with {0,1,2,3,4,5,6}

    Let's find, how many of them end with 1
    H = 5 {2,3,4,5,6}
    T = 5 {0 and the rest of {0,2,3,4,5,6} without digit taken for H}
    U = 1 {1}
    So, 5*5 = 25 numbers end with 1
    25*3 = 75 numbers end with {1,3,5}

    Among our NP(6) there are
    180 - 75 = 105 even numbers
     
  10. Jul 4, 2016 #9
    To generalize again
    In the general case there are (n-1)2 numbers that end with the chosen digit from {1...n} and n(n-1) numbers that end with 0.
    The total sum
    (n-1)2n + n(n-1) = n2(n-1)
    that confirms the formula above.
     
  11. Jul 4, 2016 #10

    Ray Vickson

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    It seems like you are doing some double counting. Some of the possibilities when you are allowed to use 0 do not actually use 0, so it does not look right to just add the 12 and the 6 possibilities.

    Better, I think, is to enumerate (1) all the possibilities that do not use 0 anywhere; and (2) the possibilities that specifically do use 0 in one of the two allowed places.
     
  12. Jul 4, 2016 #11

    SammyS

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    I did a little editing of the following to reflect and earlier suggestion I made regarding notation. Consider three digit numbers of the form "h t u" where h is the hundred's digit, t is the ten's digit, and u is the unit's digit . And as previously H is the number of possible choices for the hundred's digit, similarly for T and U .
    You have the same difficulty that you ran into in the OP.

    If you proceed from the right (What you do here with the unit's digit is fine.), Then for u = 2, 4, or 6, yes, T = 6, but you can't say that H = 4. For those cases (three of them altogether) in which t = 0, you have H = 5 .

    You can do the following for each of u = 2, 4, or 6: First choose h. There are 5 choices (can't have zero), i.e. H = 5. Then for t (now we can have t = 0), we also get T = 5.. Each gives 25 possibilities.

    Yes, we have 30 + 3×25 = 105.

    @ Ray ,
    Dr. Kaminkowa has accounted for all relevant possibilities and no more.
     
  13. Jul 4, 2016 #12

    Ray Vickson

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    I was responding to post #3; basically, the final answer was OK but the arguments to get there seemed suspect to me, or at least, were incomplete.
     
  14. Jul 4, 2016 #13

    SammyS

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    Oh, yes. I do see a problem with that counting scheme.

    However, the total number of possibilities is correct.
     
    Last edited: Jul 4, 2016
  15. Jul 4, 2016 #14

    Ray Vickson

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    Exactly: the difference is between writing 12 as 4x3 or as 3x2 + 3x2 and interpreting accordingly.
     
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