Combination problem (with exclude candidate)

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Discussion Overview

The discussion revolves around combinatorial problems involving the selection of committees under specific constraints, particularly focusing on scenarios where certain individuals cannot be included together. The participants explore different methods to calculate the number of valid combinations and express concerns regarding discrepancies between their calculations and the answers provided in a textbook.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents two methods for calculating the number of ways to form a committee of five from 11 people, considering the restriction that two specific individuals cannot be on the same committee.
  • The first method involves splitting the group into three parts and calculating combinations based on the inclusion of one of the two individuals or neither.
  • The second method calculates total combinations and subtracts those that include both restricted individuals.
  • Another participant asserts that the book's answer is incorrect and suggests that an underestimate can be achieved by excluding one of the two individuals, leading to results that match the book's answers.
  • Some participants express confusion about how the book arrived at its answers, with one confirming the calculations presented by the original poster.
  • There is speculation that the book's author may have assumed one of the two individuals must be included in the committee, which could lead to the provided answers.
  • Another participant discusses how the book's answers could be derived by ignoring combinations that include one of the two individuals or by treating the two individuals as a single case.

Areas of Agreement / Disagreement

Participants generally disagree on the correctness of the book's answers, with multiple competing views on how to approach the problem and calculate the combinations. The discussion remains unresolved regarding the validity of the book's solutions.

Contextual Notes

Participants express uncertainty about the assumptions underlying the problem and the definitions used in the calculations. There are unresolved mathematical steps in the reasoning presented.

tony24810
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In how many ways can a committee of five be chosen from 11 people, if two particular people will not work on the same committee (that is, if one is included, the other must be excluded).

I have attempted this questions with differents approaches and yielded the same answers, but my answer disagree with the answer from the book.

I would really appreciate if someone will point out if there're flaws in my arguments or if the book is wrong.method 1:

suppose the group of 11 is split into 3 groups: guy1, guy2 and the rest(9 persons)

number of ways to form committe with guy1 = 1 * 9C4
number of ways to form committe with guy2 = 1 * 9C4
number of ways to form committe from 'the rest' = 9C5

number of ways = 9C4 + 9C4 + 9C5 = 126+126+126=378

method 2:

all combinations - the combinations that contain both guy1 and guy2
= 11C5 - 9C3
= 462 - 84
=378
The answer from the book is 252.

Please someone help! thanks!the next question is similar:

In how many ways may a team of four oarsmen be chosen from a panel of 13 people if two individuals refuse to work on the same rowing team as each other?

my solution = ways with guy1 + ways with guy2 + ways without those 2
= 11C3 + 11C3 + 11C4
= 165 + 165 + 330
= 660but the book says 495
 
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The book is definitely wrong. In each case, you can make what is obviously an underestimate by simply eliminating one of the protagonists, leaving 10C5 in the first and 12C4 in the second. Intriguingly, these produce the book answers.
 
I can confirm your numbers, and have no idea how the book got its numbers.

Edit: I really checked this for 6 minutes?
 
Thanks a lot! I was starting to wonder if I should re do a statistic course lol.
 
Perhaps whoever answered the question to the book assumed that one of the two people must be on the committee.
 
This gives the book's answer of 252 for the first problem, but the answer to the second problem would be 330 in this case (2*(11 choose 3)).

You get the numbers from the book if you ignore every combination with guy 2:
252=(10 choose 5)
495= (12 choose 4)

Or, similar, if you consider (guy1, n others) and (guy2, n others) as one case (why??).
 

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