Pine Lane Middle School's student council has 10 seventh graders and 8 eighth graders serving as representatives. A subcommittee containing 3 seventh graders and 3 eighth graders is formed to organize a student dance. How many different subcommittees can be formed?

I thought about first just splitting it into groups but it proved way to inefective. For example: I would first split the eighth graders and the seventh graders and count how many ways. But it didnt work out. So can anyone please give me a hint on how to do this? TY!

AltShaka Apr 2, 2020

#1**+1 **

I'm not exaclty sure how to do this but: 2 seventh grader sub committees but put 2 extra seventh graders in each group.

Then for the eighth graders: make 2 subcommittees and put one of the extra eight graders in each subcommittee.

You may find my "stradegy" a bit confusing, so ask me if you need me to explain it.

uboachan Apr 2, 2020

#2**+1 **

Wait so i think what you mean is that you want me to have 5 seventh graders and 5 eigth graders?

AltShaka Apr 2, 2020

#4**+1 **

No, each subcommitee contains 3 seventh or eighth graders, so for seventh graders there would be 2 subcommittees, so a total of 6 seventh graders, but there are 4 kids left out, so you would put 2 extra in each subcommitee.

For the eighth graders, you would put 2 subcommittees and put 1 extra kid in each group.

uboachan
Apr 2, 2020

#3**+3 **

*Hello Altshaka!*

Total subcommittees = Seventh Grade Group * Eighth Grade Group

Seventh Grade Group = choosing 3 seventh graders out of 10.

\({10 \choose 3}=\frac{10!}{7!3!}=120\)

Eigth Grade Group = choosing 3 eighth graders out of 8

\({8 \choose 3}=\frac{8!}{5!3!}=56\)

Total subcomittees = 120 * 56 = 6720

- I'm not so sure if this is correct, but this is my attempt on the problem.

AnExtremelyLongName Apr 2, 2020

#6**0 **

You are correct! Ty very much! I like your strategy I have never thought of it before! But why do we do 7!*3!? Could you please explain? Ty!

AltShaka Apr 2, 2020

#8**0 **

You would do that since you are choosing 3 seventh graders out of ten, leaving seven remaining.

You're in sixth grade? WOW.

CalTheGreat Apr 2, 2020

#11**+1 **

Ur in elementery school according to u **But Im not so sure ill get to the bottom of this!**

AltShaka
Apr 2, 2020

#13**0 **

Why can't you believe me?! I really AM in elementary school, and that's honestly nothing to brag about...

CalTheGreat
Apr 2, 2020

#20**+1 **

No worries! I didn't even **start** these types of math problems until the **end** of 7th grade.

AnExtremelyLongName
Apr 2, 2020

#10**+3 **

We want to choose 3 of the 10 seventh graders and 3 of the 8 eighth graders

Number of possible subcommittes =

C( 10, 3) * C (8, 3) =

120 * 56 =

6720

CPhill Apr 2, 2020

#12

#18**+2 **

The reason why I have 7! and 3! is because it is part of the **combination** formula.

For example: \({10 \choose 3}=\frac{10!}{7!3!}=\frac{10*9*8*7*6*5*4*3*2*1}{(7*6*5*4*3*2*1)(*3*2*1)}\)

You are \(r\) items out of \(n\) things, then it would be \({n \choose r}\)

In this case, we are choosing 3 items out of 10 things.

AnExtremelyLongName
Apr 2, 2020