+422 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! 
+53 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. 
+422 Wait so i think what you mean is that you want me to have 5 seventh graders and 5 eigth graders?
+53 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.
+658 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.
+422 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!
+2094 You would do that since you are choosing 3 seventh graders out of ten, leaving seven remaining.
You're in sixth grade? WOW.
+422 Ur in elementery school according to u But Im not so sure ill get to the bottom of this!
+2094 Why can't you believe me?! I really AM in elementary school, and that's honestly nothing to brag about...
+658 No worries! I didn't even start these types of math problems until the end of 7th grade.
+128475 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
+658 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.
