Moonstrucked

Thank you for the answers @twohundredping I see better now where my "double count" may hide. What bugs me still, is that my double counting produced a prime number of additional outcomes (53). My guess now is that I perform a multiple count of permutations in each group of 3 biogs, and in a group of 4 biogs. So I should substract 6*5 outcomes for the sets of 3, and 23 outcomes for the set of 4 from my answer to get the right answer. @extraordinaire: As it says "2 or more biographies" I can still chose an additional biog. at a second step. The problem is though, that when I'm "trating them as equal" I double count some selections of biographies that are just permutations, as mentioned above.

As I've mentioned above, I clearly understand this solution provided by you and ETS. And I understand that it's correct. Yet I can't see where my solution (first chose two mandatory biogs, then treat ALL the books left as equal and chose any two of 8 = 2 biogs and 6 novels) is wrong. In my experience there are always several ways to count combinations, and all the right ways lead to the same answer. So it bugs me that two solutions that both seem right to me lead to different answers. And as I know that on the test day I'll come up with _my_ way of solving this kind of problem - I want to know where this way failed.

The question is from the official Revised GRE guide. Although I understand the solution provided in the book, I don't see where my logic is failing, and why my answer doesn't match the official one. Question: A reading list for a humanities course consists of 10 books, of which 4 are biographies and the rest are novels. Each student is required to read a selection of 4 books from the list, including 2 or more biographies. How many selections of 4 books satisfy the requirements? ETS Logic: Consider three cases separately: one choses 4 biographies (1), one choses 3 biographies and 1 novel (4*6=24), and one choses 2 biographis and 2 novels (4C2*6C2=90). Total number of selections would be case1+case2+case3=115 My logic: One should first chose two biographies out of 4 (4C2), and then chose any 2 books from the repaining pile (*8C2). With the total number of selections=step1*step2=168. What am I doing wrong? My guess is that I should account for some repetitions, but I can't figure out what kind of repetitions and what is the number?

Hi, me again... The answer should be D. Since n is a multiple of tree prime numbers (two of them are the same), and at the same time a multiple of 5 - either p or q =5. That makes p^2 or q^2 is a multiple of 25...

Judging by the prep-books and prep-tests I've seen so far - YES, you should know these. (And yes, it is frustrating if you're used to the good old metric system, that never deals with 12 smth in smth else etc. :))

I'd say the answer is A: Take a partial sum of the series or regroup to see a pattern. Here S = (1 - 1/2) + (1/3 - 1/4) + ... First term is 1/2, each subsequent term is >0, hence the sum S is >1/2. Hope that helps :)

Hi, Avi I've been trying to solve this problem (today ;)). Unfortunately the one I found has a russian interface. But my general advise would be to look for a basic word-learning program that would allow you to add your own dictionary. It's easier to find free software without 'GRE' label on it than with it... (The program I found is here: BX Language acquisition and it also doesn't have GRE words by default but allows you to add stuff you want to learn)