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# Help?

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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!

Apr 2, 2020

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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.

Apr 2, 2020
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Wait so i think what you mean is that you want me to have 5 seventh graders and 5 eigth graders?

Apr 2, 2020
#4
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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
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Ah shoot, i just realized my answer wouldn't make sense.

uboachan  Apr 2, 2020
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Hello Altshaka!

$${10 \choose 3}=\frac{10!}{7!3!}=120$$

$${8 \choose 3}=\frac{8!}{5!3!}=56$$

Total subcomittees = 120 * 56 = 6720

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

Apr 2, 2020
#5
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You might be right. I'm just a clueless eighth grader looking to help people.

uboachan  Apr 2, 2020
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Wait why does 10/3 = 10/7!*3!.? Shouldnt 10/7!*3! equal to 10/30240?

AltShaka  Apr 2, 2020
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Good job, AELN   !!!!!

Key - Wrecked    !!!!!

CPhill  Apr 2, 2020
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Cphill i agree! But why does !7*!3 = 3 in his thing?

AltShaka  Apr 2, 2020
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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!

Apr 2, 2020
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Well you can do it! Im in 6th grade and i TRY  to help people. (not always 100%)

AltShaka  Apr 2, 2020
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You would do that since you are choosing 3 seventh graders out of ten, leaving seven remaining.

Apr 2, 2020
edited by CalTheGreat  Apr 2, 2020
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Ur in elementery school according to u But Im not so sure ill get to the bottom of this!

AltShaka  Apr 2, 2020
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Why can't you believe me?! I really AM in elementary school, and that's honestly nothing to brag about...

CalTheGreat  Apr 2, 2020
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Ha! Really smart for one! Unlike me....

AltShaka  Apr 2, 2020
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No worries! I didn't even start these types of math problems until the end of 7th grade.

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I have other external clasess so yah

AltShaka  Apr 2, 2020
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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

Apr 2, 2020
#12
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Wow you all have good ways of thinking! I never thought of that.

Apr 2, 2020
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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.

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Ohh  thank you! I learend that today because of you!!

Apr 2, 2020