Elementary combinatorics problem - why am I wrong?

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    Combinatorics Elementary
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Discussion Overview

The discussion revolves around a combinatorics problem involving the selection of 3-person groups from a mix of experienced and inexperienced individuals, specifically addressing the counting methods used to arrive at the correct number of combinations while adhering to the condition that at most one member can be inexperienced.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant presents a calculation for the number of combinations, questioning the use of a 2-permutation versus a 3-permutation in their reasoning.
  • Another participant suggests using the binomial coefficient "6 choose 2" to determine the number of ways to select experienced individuals, along with the selection of the inexperienced member.
  • A participant seeks clarification on their misunderstanding of the counting method, specifically regarding the visual representation of their flaw.
  • Further explanation is provided about the importance of addressing double counting rather than just order, emphasizing that only the order of like objects (experienced individuals) needs to be considered.
  • A participant expresses gratitude for the clarification received, indicating a resolution to their confusion.

Areas of Agreement / Disagreement

Participants engage in a constructive dialogue to clarify the counting method, with some agreeing on the need to address double counting specifically for experienced individuals. However, the discussion does not reach a consensus on the initial misunderstanding, as it is framed as a personal flaw rather than a broader disagreement.

Contextual Notes

The discussion highlights the nuances in combinatorial reasoning, particularly in how to appropriately account for different types of individuals in a selection process. There are unresolved aspects regarding the initial participant's rationale and how it contrasts with the established combinatorial principles discussed.

Helicobacter
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6 unique experienced persons
2 unique inexperienced persons

How many 3-person combinations if at most one in the group can be inexperienced?

6*5*4/3! + 6*5*2/2! is the answer.

My answer was
6*5*4/3! + 6*5*2/3!

Why does the book divide by a 2-permutation and not a 3-permutation for the second term (all combinations consisting of 2 experienced persons and 1 inexperienced person)? My (flawed) rationale is that you have three slots to be filled; we have 6 experienced choices for the first slot, 5 remaining experienced choices for the second slot, and 2 inexperienced choices for the last slot. Since, we don't consider order, we divide by a correction factor of 3 slots permuted.
 
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If you have a group consisting of 2 experienced and one inexperienced, there are 6 choose 2 ways of picking the 2 experienced people, and 2 ways to pick the inexperienced one. Try multiplying those out
 
Thanks for your answer, but I'm looking for my flaw visually; why am I under-counting in the second term. Why is my last paragraph explanation wrong?
 
Your main goal is not to take care of order, you want to take care of double counting. Usually these things are the same, but when dividing out you only want to take care of order within each subset of like objects. So here we have one pair of like objects (the experienced people) whose order is irrelevant, meaning you only divide by 2!

So if the inexperienced people are named Adam and Abe, and some experienced people are Bob, Billy and Bubba:

One possibility is to pick Bob, Bubba and Adam

One possibility is to pick Bubba, Bob and Adam. (so the group of Bob, Bubba and Adam has been counted twice)

Picking Adam, then Bubba then Bob is not a possibility because you said you were picking two experienced people, then one inexperienced person, so you never counted this possibility to begin with. (so the group was only counted twice, not more times)
 
That finally cleared it up for me. Thanks a lot.
 

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