Permutations of the letters a, b, c, d, e, f, g

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Discussion Overview

The discussion revolves around calculating the number of permutations of the letters a, b, c, d, e, f, g with specific conditions on the spacing between the letters a and b. The focus is on combinatorial reasoning and mathematical calculations related to permutations.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant proposes that the number of permutations with two letters between a and b can be calculated as 5! and similarly for three letters, suggesting a total of 5! + 5! as the answer.
  • Another participant agrees with the initial calculation but emphasizes the need to consider all possible position pairs for a and b, indicating that the initial approach is incomplete.
  • A later reply introduces a revised calculation of (5! * 4) + (5! * 3) and seeks confirmation on its correctness.
  • Another participant points out a missing factor of 2 in the revised calculation, noting that the order of a and b matters unless a is assumed to always precede b.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the final answer, as there are differing views on the factors to include in the calculations and the assumptions regarding the order of a and b.

Contextual Notes

There are unresolved assumptions regarding the positioning of a and b, and the implications of their order on the total count of permutations.

Caldus
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How many permutations of the letters a, b, c, d, e, f, g have either two or three letters between a and b?

My guess for this is that if a and b have to have two letters between them, then there are 5! ways to arrange the rest of the letters, right? Same deal if a and b have to have three letters between them. So:

5! + 5!

Would be my answer. Am I anywhere close with this one? Thank you.
 
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Your answer is correct for each specific (a,b) position pair. Your have to multiply by all possible position pairs for a and b, e.g. (1,4), (4,1), (2,5) etc. for two spacing, and similarly for three spacing.
 
Thank you.

I came up with this new answer:

(5! * 4) + (5! * 3) = 840

Am I close here?
 
You're missing a factor of 2, since a at 1 and b at 4 is different from b at 1 and a at 4, etc., unless you are assuming a is always before b. In that case you are right.
 
Last edited:

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