Counting problem (book wrong?)

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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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  • #2
Dick
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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.
The book's solution is correct. Their illustration of the cases is stupidly wrong. There are four cases. What's the missing one?
 
  • #3
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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?
Case 2:
9 d 9 9.
 
  • #4
Dick
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Case 2:
9 d 9 9.
Sure. There are four cases, not three. That's a pretty odd mistake to make.
 
  • #5
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Sure. There are four cases, not three. That's a pretty odd mistake to make.
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.
 
  • #6
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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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  • #7
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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....?
 
  • #8
Dick
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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....?
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?
 
  • #9
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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?
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
 
  • #10
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Nevermind I see where I went wrong.
 
  • #11
Dick
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Nevermind I see where I went wrong.
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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