4-member committee from 4 grades with 2 reps each

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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: .[tex]_8C_4 \,=\,70[/tex] possible committees.

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

Therefore, there are: .[tex]70 - 15 \,=\,55[/tex] 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.
. . [tex]_6C_2 \,=\,15[/tex] ways.

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

Therefore, there are: .[tex]15 + 15 \,=\,30[/tex] 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?