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Counting problem (book wrong?)

  1. Nov 24, 2013 #1
    Question and my solution is in the paint document.

    Case 1 - Case 3 represents all possible combinations of string of length 4 with exactly three 9's.

    The d represents a digit that is not 9.
    d = 0 or 1 or 2 or 3 or ... or 8: This is 9 options.

    Notice that none of the cases 1 - 3 have identical strings because,
    Case 1 cannot be equal to any of the strings in case 2 or 3 because Case 1 start with a 9 where case 2 starts with d which cannot be a 9 also case 1 ends with a 9 where case 3 ends with a d which cannot be a 9. Finally the third column of case 2 is a d and the third column of case 3 is a 9.
    Therefore each case has 9 distinct strings with exactly three 9's and 9(3) = 27.
    The book solution is
    9 + 9 + 9 + 9 = 36.
    Book solution is in paint doc.
     

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  3. Nov 24, 2013 #2

    Dick

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    The book's solution is correct. Their illustration of the cases is stupidly wrong. There are four cases. What's the missing one?
     
  4. Nov 24, 2013 #3
    Case 2:
    9 d 9 9.
     
  5. Nov 24, 2013 #4

    Dick

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    Sure. There are four cases, not three. That's a pretty odd mistake to make.
     
  6. Nov 24, 2013 #5
    Sure is, thanks for your help. Im going to post another one in this thread in a minute maybe you can help me with this one.
     
  7. Nov 24, 2013 #6
    Case 1: Start with BO (In that order)
    Case 2: End with BO (In that order)
    Case 3: Start with BO and end with BO (In that order)

    L represents All letters not B, therefore all strings in a case are different from all other strings in the other two cases.

    Case 1: Start with BO, therefore column 3 - 6 and 8 have 26 options column 7 has 25 options,
    265*25 total options.

    Case 2: 265*25 total options.

    Case 3: 264 Total options

    Sum of all options from Case 1-3 is 594525776.

    The answer is 617374576.
    I am way off.

    I have a second solution in post number 7 (which is correct) but I want to know why the solution in this post is wrong.
     

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  8. Nov 24, 2013 #7
    However I get the correct answer if I have the two following cases.

    Case 1: B O _ _ _ _ _ _
    26^6 total options
    Case 2: _ _ _ _ _ _ B O
    26^6 total options

    Now if I look at Case 2 and compare it to case 1 the only strings that are identical would be the strings of the form,

    B O _ _ _ _ B O
    26^4 options that are in common.
    Therefore the answer is

    26^6 + 26^6 - 26^4.

    I just want to know why my first attempt failed....?
     
  9. Nov 24, 2013 #8

    Dick

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    Your second attempt is so clearly correct, I'm having trouble trying to figure out why you think the first attempt might be correct. What makes column 7 different from column 8? The number that start with BO and don't end with BO is 26^4(26^2-1), isn't it?
     
  10. Nov 24, 2013 #9
    Case 1: Col 7 is different from col 7 of cases 2 and 3 because L represents all letters that are not B.

    The number 26^4(26^2-1) is referring to case 1 correct? The reason I have 25*26^5 is because col 1 and col 2 are set therefore my only options are col 3,4,5,6, and 8 which can all be letters a through z, 26 options. col 7 can be any letter that is not B therefore there are 25 options.
    That is how I arrived at the number 25*26^5
     
  11. Nov 24, 2013 #10
    Nevermind I see where I went wrong.
     
  12. Nov 24, 2013 #11

    Dick

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    I hope so. For your case 1, letters 3-6 can be anything. Letters 7-8 can be anything except BO. That's 26^4*(26^2-1).
     
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