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

[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?

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.

 

[soroban]

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.

 

[Raerin]

Also, what if both (b and c) restrictions apply? Would it be 55+30? The answer says it's 23.

 

[I like Serena]

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?

 

[CellCoree]

a) Your answer of 70 is correct. The formula for choosing a committee of 4 members from a group of 8 (4 girls and 4 boys) is 8C4 = 8!/(4!4!) = 70.

b) Your answer of 40 is correct. The formula for choosing a committee of 3 members from a group of 6 (3 girls and 3 boys) is 6C3 = 6!/(3!3!) = 20. However, since David and Ryan cannot both be on the committee, we need to subtract the number of committees where they are both chosen. This can be calculated as 2C1 = 2. Therefore, the total number of committees is 6C3 * 2C1 = 40 - 2 = 38.

c) For this restriction, we need to consider two cases: Leigh and Sarah are both on the committee, or they are both off the committee.

Case 1: Leigh and Sarah are both on the committee. In this case, we need to choose 2 more members from the remaining 6 students (2 girls and 4 boys). This can be done in 6C2 = 6!/(2!4!) = 15 ways.

Case 2: Leigh and Sarah are both off the committee. In this case, we need to choose 4 members from the remaining 6 students (3 girls and 3 boys). This can be done in 6C4 = 6!/(4!2!) = 15 ways.

Therefore, the total number of committees is 15 + 15 = 30.