1. R
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    11 Mar '21 03:47
    How many 4 digit numbers contain not more that two different digits?
  2. Standard memberBigDogg
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    11 Mar '21 07:59
  3. R
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    11 Mar '21 12:23
    Big Dogg is Correct!
  4. Subscribervenda
    Dave
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    11 Mar '21 14:49

    Removed by poster

  5. Subscribervenda
    Dave
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    11 Mar '21 19:22
    Once again,I have successfully arrived at the wrong answer!
    I realised my mistake on the post I deleted.
    However I still have the wrong answer according to the answer given by dogg.
    I would appreciate if someone could explain(preferably in language a 5 yr old can understand)what is wrong with my logic:-
    The first digit selected cannot be a zero, so the number of digits available for selection is 9
    So F(1st = 9)
    The second digit selected if different from F is also 9
    So S = 9
    The third digit selected can only be one of 2 (F or S)
    So T =2
    The 4th digit can only be 1(either the same as F or the same as S)
    There are 9 possibilities with only 1 digit so the sum is 9*9*2 +9 =171
    OR the first selection is one of 9
    The second selection is [b]the same as[b/] the first =1
    The third selection is again 1 of the 9 remaining digits
    The 4th selection must be the same as the third which gives the sum 9*1*9*1 +
    9 = 90 so both must be wrong.
    Once you get this train of thought in your head it is difficult to dislodge.
    Sorry for appearing to be so thick!!
  6. Standard memberBigDogg
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    11 Mar '21 21:29
    @venda said
    Once again,I have successfully arrived at the wrong answer!
    I realised my mistake on the post I deleted.
    However I still have the wrong answer according to the answer given by dogg.
    I would appreciate if someone could explain(preferably in language a 5 yr old can understand)what is wrong with my logic:-
    The first digit selected cannot be a zero, so the number of digits ava ...[text shortened]... his train of thought in your head it is difficult to dislodge.
    Sorry for appearing to be so thick!!
    That is an excellent line of thought, actually. It just needs some refinement.

    Let's call the four digits F, S, T, and H, respectively.

    Case 1: F is not equal to S

    As you said, there are 9 possible digits for F.
    Combine that with another 9 possible digits for S.
    T must equal F, or S, so there are 2 possible digits for T.
    However, there are also two possible digits for H! It, like T, can equal either F, or S.

    9*9*2*2 = 324 possibilities

    Case 2: F is equal to S

    This one must be broken down into two sub-cases.

    Sub-case A: F == S == T

    F has 9 possibilities. S and T only have 1.
    H has 10 different possibilities. It can be any of 0 through 9.

    9*10 = 90 possibilities

    Sub-case B: T is not equal to F, and not equal to S

    9 possibilities for F. 1 for S.
    There are 9 possibilities for T (any of 0 through 9, except for F).
    There are 2 possibilities for H. (either F, or T)

    9*9*2 = 162 possibilities

    Total: 324 + 90 + 162 = 576 possibilities.
  7. Subscribervenda
    Dave
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    12 Mar '21 09:21
    @bigdoggproblem said
    That is an excellent line of thought, actually. It just needs some refinement.

    Let's call the four digits F, S, T, and H, respectively.

    Case 1: F is not equal to S

    As you said, there are 9 possible digits for F.
    Combine that with another 9 possible digits for S.
    T must equal F, or S, so there are 2 possible digits for T.
    However, there are also two p ...[text shortened]... s for H. (either F, or T)

    9*9*2 = 162 possibilities

    Total: 324 + 90 + 162 = 576 possibilities.
    Thanks for the explanation.Makes it a lot clearer
    Seems I am like the horse in the steeplechase that leads all the way and then stumbles and falls at the last hurdle!
  8. Standard memberBigDogg
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    13 Mar '21 00:38
    @venda said
    Thanks for the explanation.Makes it a lot clearer
    Seems I am like the horse in the steeplechase that leads all the way and then stumbles and falls at the last hurdle!
    Truth be told, I wrote a program to solve, first. Figured out the math later, again using programming to help.
  9. Subscribervenda
    Dave
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    13 Mar '21 09:55
    @bigdoggproblem said
    Truth be told, I wrote a program to solve, first. Figured out the math later, again using programming to help.
    I can do most things with spreadsheets but I've never done any programming as such apart from the odd macro when I was at work.
    As an aside, I downloaded a script for my chromebook to do a task for me.
    I found out later I didn't really need it but can I get rid of it?!!
    I've tried everything except going back to factory settings which I don't want to do.
    Every time I open any of my spreadsheets now I get the message "working" and it won't respond for a few seconds.
    These things are sent to try us.
  10. R
    Standard memberRemoved
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    13 Mar '21 20:55
    Here is how I did it.


    Type 0:

    Single digit used - 4 digit numbers: aaaa

    There are 9 choices for the digit.

    Type 0 = 9

    Two digits used - 4 digit numbers:

    Type 1: aabb

    There are C(4,2) = 6 ways to arrange 2 a's and 2 b's

    aabb
    bbaa

    abab
    baba

    abba
    baab

    However, notice that due to symmetry the numbers in this Type are represented 2 fold: aabb = bbaa ; abab = baba ; and abba = baab

    So divide by 2.

    In each of the 3 remaining subgroups there are 9 ways to choose the leading digit and 9 ways remaining digit

    Type 1 = 9*9*3 = 243

    Type 2: aaab

    There are C(4,3) = 4 ways to arrange 3 a's and 1 b

    We don't have the symmetry to worry about in this type and there are 9 ways to choose the leading digit and 9 ways remaining digit.

    Type 2 = 9*9*4 = 324

    Thus the total set of 4 digit numbers that use not more that 2 distinct digits:

    Type 0 + Type 1 + Type 2 = 9 + 243 + 324 = 576
  11. Standard memberBigDogg
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    14 Mar '21 06:08
    @venda said
    As an aside, I downloaded a script for my chromebook to do a task for me.
    I found out later I didn't really need it but can I get rid of it?!!
    You'd have to find the location where you downloaded it, and delete it from there, or find out where you installed it, and uninstall from there.

    If you cannot find it, try googling "uninstall chromebook script".
  12. Subscribervenda
    Dave
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    14 Mar '21 10:18
    @bigdoggproblem said
    You'd have to find the location where you downloaded it, and delete it from there, or find out where you installed it, and uninstall from there.

    If you cannot find it, try googling "uninstall chromebook script".
    Thanks.
    I've done all that before and the script doesn't work anymore but it still comes up with the "working" message .
    There must be something it's put in the rom or perhaps it's nothing to do with the script at all!
    Not a big issue,just a little annoying.
    Otherwise, I can recommend chromebooks.
    There are things that are unsupported but most features are very similar to microsoft without the need for security and windows updates that take ages.When I get "update available restart to install" message I click on it and the chromebook re boots in less than 15 seconds.
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