<p>The student committee to investigate the missing prom funds is to be made up of four members of the student council. If there are seven members in the student council, then how many different committees are possible?</p>
<p>I thought the answer would be 7 nCr 4 = 35, but it's apparently 7 nPr = 840. Why? A committee position is a committee position, it's not like they're numbered Position 1 Position 2 etc.</p>
<p>It depends - if all the positions are considered the same it’s 7 nCr 4, however if you consider them to be four separate positions then you have 7 nPr 4. So perhaps it was a poorly worded question - which book was it from?</p>
<p>I guess its because once a member is in one committee then he is unable to be in another.</p>
<p>^ Really? (10char)</p>
<p>No nvm, I really have no idea why I said that. The question was probably worded badly. I think the question was however trying to imply that there are distinct positions within the committee</p>
<p>Badly worded question, but I’d say this is a combination problem, which would make the answer 35.</p>
<p>I could be wrong but I think the answer is 7 choose 4, which equals 210.</p>
<p>It’s from Sparknotes, which I’m only using because I need to brush up on permutations/combinations. </p>
<p>I just wanted to make sure I wasn’t going crazy or anything.</p>
<p>4 spots available for 7 people. so you have four open spots. for the first spot there are 7 avaliable candidates, for the second (since the first one is take) there are only 6 available candidates and so on</p>
<p>so 7<em>6</em>5*4= 800</p>
<p>Jamesford – you had it right from the start. It is 7C4 = 35. </p>
<p>easyasabc – you are doing 7P4. but the problem calls for making a group, but not giving distinctive positions within the group. So for example, ABCD and ABDC and BADC etc… are all the same committee. </p>
<p>Also, this is yet another example of how you can become frustrated working with material that is not from the college board. I’m not saying they are all bad, but they are sporadically bad in ways that are annoying, causing you to work on material that is never on the test or in this case to doubt yourself unneccessarily.</p>
<p>Yeah, whenever you hear “committee” problems, they’re usually combination.</p>
<p>My math is rusty, 7 choose 4 equals 35.</p>
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<p>I never use non CB practice tests, but this is a concept in particular that I don’t recall (and I don’t feel like riffling through a bunch of tests just to find one problem).</p>