In how many ways can a committee be formed....

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Discussion Overview

The discussion revolves around the combinatorial problem of forming a 4-member committee from representatives of grades 9 to 12, considering various restrictions on membership. Participants explore different scenarios, including the absence of restrictions, specific individuals not being allowed to serve together, and conditions regarding the inclusion of certain members.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants calculate the total number of committees without restrictions as 8C4 = 70, while others assert the correct answer is 28, leading to disagreement.
  • For the scenario where David and Ryan cannot both be on the committee, one participant calculates 6C3 * 2C1 = 40, while another claims the correct answer is 55, suggesting a different approach to the problem.
  • In the case where Leigh and Sarah must either be on or off the committee, one participant proposes that if both are included, 2 more members can be chosen from the remaining 6, resulting in 6C2 = 15 ways, and if neither is included, all 4 must come from the other 6, yielding 6C4 = 15 ways, totaling 30 ways.
  • Another participant questions how to combine the restrictions from parts (b) and (c), suggesting that the total might be 55 + 30, but notes that the answer is stated to be 23, indicating confusion and uncertainty about the calculations.
  • Further exploration of combinations under the restrictions is proposed, with various cases outlined for how David, Ryan, Leigh, and Sarah might be included or excluded from the committee.

Areas of Agreement / Disagreement

Participants express differing views on the correct answers to the committee formation scenarios, with no consensus reached on the total number of ways to form the committee under the specified restrictions.

Contextual Notes

Participants' calculations depend on the interpretation of the restrictions and the combinatorial methods applied, with unresolved assumptions regarding the definitions of the scenarios presented.

Raerin
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In how many ways can a 4-member committee be formed from a girl rep and a boy rep from each of grades 9, 10, 11, and 12?

a) There is no restriction?
I did 8C4 = 70, but apparently the correct answer is 28. I just want to confirm if my answer is right or not.

b) David and Ryan can not both be on the committee?
6C3 * 2C1 = 40 but, again, the apparent answer is 55.

c) Leigh and Sarah will either be on or off the committee?
I have no idea what to do for this restriction.
 
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Hello, Raerin!

In how many ways can a 4-member committee
be formed from a girl rep and a boy rep from
each of grades 9, 10, 11, and 12 if:

(a) There is no restriction?
I did 8C4 = 70, but apparently the correct answer is 28.
I just want to confirm if my answer is right or not.
You are right . . . "They" are wrong.

This is evident in their answer to part (b).
(b) David and Ryan can not both be on the committee?
6C3 * 2C1 = 40 but, again, the apparent answer is 55.
Your answer is incorrect; they are right.

There are: ._8C_4 \,=\,70 possible committees.

If David and Ryan are both on the committee,
we must choose the other 2 from the other 6 people.
. . there are: _6C_2 = 15 ways.

Therefore, there are: .70 - 15 \,=\,55 ways
. . in which David and Ryan are not serving together.


*snicker*

(a) They claim there are 28 possible committees.
(b) They say 55 of them do not have both David and Ryan.

I guess it's true:
. . Five out of four people have trouble with marh.
(c) Leigh and Sarah will either be on or off the committee?
I have no idea what to do for this restriction.
If Leigh and Sarah are both on the committee,
choose 2 more from the other 6 people.
. . _6C_2 \,=\,15 ways.

If neither Leigh and Sarah are on the committee,
choose all 4 from the other 6 people.
. . _6C_4 \,=\,15 ways.

Therefore, there are: .15 + 15 \,=\,30 ways.
 
Also, what if both (b and c) restrictions apply? Would it be 55+30? The answer says it's 23.
 
Raerin said:
Also, what if both (b and c) restrictions apply? Would it be 55+30? The answer says it's 23.

With more restrictions the number can only go down...

Possibilities are:
  1. David on, Ryan not, Leigh & Sarah on
  2. David not on, Ryan on, Leigh & Sarah on
  3. Neither David nor Ryan on, Leigh & Sarah on
  4. David on, Ryan not, Leigh & Sarah not on
  5. David not on, Ryan on, Leigh & Sarah not on
  6. Neither David nor Ryan on, Leigh & Sarah not on
How many combinations in each case?
 

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