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Mathematics 22 Online
OpenStudy (anonymous):

Help!?!? Please??? In how many ways can a committee consisting of 4 faculty members and 5 students be formed if there are 12 faculty members and 14 students eligible to serve on the committee?

OpenStudy (kropot72):

The number of ways is 12C4 + 14C5

OpenStudy (shubhamsrg):

they should be multiplied since they are happening simultaneously.

OpenStudy (anonymous):

so 12*4 and 14*5?

OpenStudy (anonymous):

12!/8!4!*14!/9!5!

OpenStudy (kropot72):

@shubhamsrg You are right. Each combination of faculty members can be taken with each combination of students.

OpenStudy (shubhamsrg):

^_^ one of the rare achievements for me! :|

OpenStudy (kropot72):

@maddiemaze Do you understand how to calculate 12C4 * 14C5 ?

OpenStudy (anonymous):

@shubhamsrg Modest hmm..

OpenStudy (shubhamsrg):

look who is saying! -_-

OpenStudy (anonymous):

no I don't

OpenStudy (kropot72):

The method of finding the number of combinations of 12 different things taken 4 at a time comes from the general formula for n things taken r at a time , which is\[nCr=\frac{n!}{(n-r)!r!}\] \[12C4=\frac{12!}{(12-4)!4!}=\frac{12\times 11\times 10\times 9}{4\times 3\times 2\times 1}\] Can you try finding 14C5 ?

OpenStudy (anonymous):

I think it would be 2002

OpenStudy (kropot72):

Correct! Good work. So what do you get when you multiply the two numbers of combinations together?

OpenStudy (anonymous):

990990

OpenStudy (kropot72):

Absolutely correct! Well done.

OpenStudy (anonymous):

Thanks for your help

OpenStudy (kropot72):

You're welcome. Sorry for my bad at the start :(

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