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Mathematics 17 Online
OpenStudy (anonymous):

9 students volunteer for a committee. How many different 2-person committees can be chosen?

OpenStudy (anonymous):

so you know from the problem you can have 2 students in a group

OpenStudy (anonymous):

if you divide 9 /2, you get a fraction but we can't have a fraction of a group so take the nearest whole number

OpenStudy (mathstudent55):

In this case, order is not important. Let's say the students are named A through I. Having A and B in a committee is the same as having B and A in the committee.

OpenStudy (anonymous):

I Got 36 ?

OpenStudy (anonymous):

I wonder as I went through that logic if this problem is a little more complicated than it appears to be

OpenStudy (anonymous):

The answer choses are a.362,880 b.1 c.72 d.36

OpenStudy (anonymous):

would you need to know how many students there are to find how many committees?

OpenStudy (mathstudent55):

You need \(_9C_2\). In general, \(_nC_r = \dfrac{n!}{(n - r)!r!}\) \( _9C_2 = \dfrac{9!}{(9 - 2)! 2!} \) \(= \dfrac{9 \times 8 \times \cancel{7!}}{\cancel{7!}\times 2} \) \(= 36\)

OpenStudy (mathstudent55):

You are correct, 36 is the answer.

OpenStudy (anonymous):

alright thank you guys !

OpenStudy (mathstudent55):

wlcm

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