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

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The discussion revolves around forming a 4-member committee from representatives of grades 9 to 12, with various restrictions. In the first scenario without restrictions, the correct number of combinations is 70, not 28 as initially thought. When excluding both David and Ryan from being on the committee together, the total ways to form the committee is 55. For the case where Leigh and Sarah are either both included or excluded, there are 30 valid combinations. The final question about applying both restrictions leads to a total of 23 valid combinations, highlighting the complexity of the problem.
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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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