Counting problem (book wrong?)

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Homework Help Overview

The discussion revolves around a combinatorial counting problem involving strings of a specified length containing certain digits or letters. Participants are examining the number of distinct strings that can be formed under specific conditions, questioning the correctness of a provided solution in a textbook.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore different cases for counting strings with specific digits and letters, questioning the completeness of the cases presented. There is a focus on identifying missing cases and understanding overlaps in string combinations.

Discussion Status

Some participants have identified discrepancies in the number of cases considered in the original problem, suggesting that there may be additional cases that need to be accounted for. Others are analyzing their own approaches to determine where their reasoning may have diverged from the expected outcomes.

Contextual Notes

There are mentions of specific constraints regarding the digits and letters used in the strings, as well as the need to consider distinct cases in the counting process. Participants are also reflecting on the implications of their assumptions regarding the arrangement of characters.

Miike012
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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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Miike012 said:
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?
 
Dick said:
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.
 
Miike012 said:
Case 2:
9 d 9 9.

Sure. There are four cases, not three. That's a pretty odd mistake to make.
 
Dick said:
Sure. There are four cases, not three. That's a pretty odd mistake to make.

Sure is, thanks for your help. I am going to post another one in this thread in a minute maybe you can help me with this one.
 
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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Last edited:
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...?
 
Miike012 said:
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?
 
Dick said:
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
Nevermind I see where I went wrong.
 
  • #11
Miike012 said:
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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