Bamse Posted March 21, 2012 Posted March 21, 2012 (edited) I haven't taken a math class in forever and it seems like the little I used to know is almost completely gone. Can anyone explain how to tackle the following problem? "Kristen must select three different items for each dinner she will serve. The items are to be chosen from among five different vegetarian and four different meat selections. If at least one of the selections must be vegetarian, how many different dinners could Kristen create?" This is the type of problem that makes me go "Who cares? I'm in the social sciences" but I guess that's not an acceptable answer. The right answer is (spoiler alert) "80". Can anyone explain to me how to solve this type of problem? I happened to get it right by guessing but I don't want to trust my luck on the real exam. Thanks! Edited March 21, 2012 by Bamse
linmich Posted March 21, 2012 Posted March 21, 2012 Seems we need to choose 3 items out of 9. But we need to rule out the case that all 3 items are meat, which is 3 out of 4 items. To choose 3 out of 9, there are 9!/(6!*3!)=84 ways. To choose 3 out of 4, there are 4!/(1!*3!)=4 ways. So in all, it is 84-4=80.
Bamse Posted March 21, 2012 Author Posted March 21, 2012 Ok thanks! Remember, you are talking to someone who doesn't know much about math so would you mind walking me through the steps? So, to check how many different combinations there are for 3 selections out of 9 total I do 9!/(6!*3!)=84 ways . Is there a formula I can memorize for this? Is it: total number of items ! / (total number of items - number of items to be selected !) x number of items to be selected ! ??
hungry Posted March 21, 2012 Posted March 21, 2012 I think this was the hardest part of the GRE, but you can learn about combinations and permutations here: http://www.mathsisfun.com/combinatorics/combinations-permutations.html or anywhere else online, there's tons of tutorials. The princeton review book for the GRE also has a great section on probability.
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