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How many different combinations are there?

  • 24-05-2018 8:39am
    #1
    NeVeR
    Registered Users, Registered Users 2 Posts: 5,356 ✭✭✭


    Ok,

    So you have the digits 0-9 and the letters A-Z

    Do many 2 digit combinations can be made ? You can use the same letters/number twice in a combo.

    Example:
    11 is ok
    AA is ok
    1A is ok
    A1 is ok

    So just because you use 1A in one combo, A1 is still a different combo and valid

    etc...etc..

    I was trying to figure out what the math's is for this but was unsure.

    We have 10 numbers/ digits and 26 letters so 36 in total

    So how many 2 digit combo's can you make ?


Comments

  • #2
    What Username Guidelines
    Registered Users, Registered Users 2 Posts: 2,139 ✭✭✭What Username Guidelines


    1,296


  • #3
    AlexisM
    Registered Users, Registered Users 2 Posts: 250 ✭✭AlexisM


    36 * 36 = 1,296 combinations


  • #4
    Your Face
    Closed Accounts Posts: 7,275 ✭✭✭Your Face


    I wish I were as drunk as the OP


  • #5
    NeVeR
    Registered Users, Registered Users 2 Posts: 5,356 ✭✭✭NeVeR


    What Username Guidelines wrote: »
    1,296

    Really.. is that all ?

    I thought there would have been more.. thanks.


  • #6
    Omackeral
    Registered Users, Registered Users 2 Posts: 32,956 ✭✭✭✭Omackeral


    NeVeR wrote: »
    math's
    NeVeR wrote: »
    combo's

    Forget the numbers and sort this stuff out. You don't have to stick in an apostrophe just because a word ends with the letter s.


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  • #7
    Peregrine
    Moderators, Education Moderators Posts: 26,403 Mod ✭✭✭✭Peregrine


    NeVeR wrote: »
    Really.. is that all ?

    I thought there would have been more.. thanks.

    Nope, that's it. 36x36 = 1296.


  • #8
    NeVeR
    Registered Users, Registered Users 2 Posts: 5,356 ✭✭✭NeVeR


    Omackeral wrote: »
    Forget the numbers and sort this stuff out. You don't have to stick in an apostrophe just because a word ends with the letter s.

    Thank's


  • #9
    Ubbquittious
    Registered Users, Registered Users 2 Posts: 6,292 ✭✭✭Ubbquittious


    Why would anyone bother using all 26 letters? Is A to F not good enough suddenly?


  • #10
    Pherekydes
    Registered Users, Registered Users 2 Posts: 16,202 ✭✭✭✭Pherekydes


    Combinatorics in AH! :)

    If we use the set of integers, how many combinations now, smartypants? :pac:


  • #11
    Hector Bellend
    Closed Accounts Posts: 3,667 ✭✭✭Hector Bellend


    yes.

    1296.

    26 + 10 = 36

    you have 36 symbols that could be matched with any of the same 36 symbols

    therefore 36 * 36=1296


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  • #12
    Hector Bellend
    Closed Accounts Posts: 3,667 ✭✭✭Hector Bellend


    Ubbquittious wrote: »
    Why would anyone bother using all 26 letters? Is A to F not good enough suddenly?

    It's only 40% of fanny.

    Therefore 26 letters is better.


  • #13
    Tokyo
    Registered Users, Registered Users 2 Posts: 22,656 ✭✭✭✭Tokyo


    There are 26 letters plus 10 digits in the pool of characters, giving us 36 characters in total - 36 choices for the first character, 36 choices for the second character (and 36 choices for the third character, 36 choices for the fourth character and so on of there were more).

    Product rule dictates that the number of combinations will be 36^n, where n is the number of digits in the combination. So for a 2 digit combination, the number of possibiliites is 36^2, or 1296 possible combinations.

    Obviously, if you want more combinations, you increase the number of digits in the combination, i.e.

    3 digit combination = 36^3 = 46656 possible combinations.
    4 digit combination = 36^4 = 1679616 possible combinations.

    ... which is pretty much why you don't get 2-digit combination locks.


  • #14
    yeppydeppy
    Registered Users, Registered Users 2 Posts: 1,163 ✭✭✭yeppydeppy


    Wouldn't 36*36 include each double twice? As in AA and AA, 11 and 11 etc? What if you only wanted to count any double once?


  • #15
    Ficheall
    Registered Users, Registered Users 2 Posts: 9,034 ✭✭✭Ficheall


    yeppydeppy wrote: »
    Wouldn't 36*36 include each double twice? As in AA and AA, 11 and 11 etc? What if you only wanted to count any double once?


    No. You have 36 options for the first column and 36 options for the second column.

    Eg.
    AA
    AB
    AC
    .
    .
    .
    ZA
    ZB
    ZC


    You don't have a second A that can go in the first column to worry about.

    If you were asking how many permutations of the word "yeppy" there were, then yes, you would have to take into account the fact that there are two "p"s.


  • #16
    Hector Bellend
    Closed Accounts Posts: 3,667 ✭✭✭Hector Bellend


    yeppydeppy wrote: »
    Wouldn't 36*36 include each double twice? As in AA and AA, 11 and 11 etc? What if you only wanted to count any double once?

    You start with 36 characters.

    If you combine AA. You are left with 35 characters on each side.

    Therefore you are left with 35 * 35 remaining possible combinations.


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