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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## A test maker is to design a probability test from a list... tagged by: BTGmoderatorLU ##### This topic has 3 expert replies and 0 member replies ### Top Member ## A test maker is to design a probability test from a list... A test maker is to design a probability test from a list of 21 questions. The 21 questions are classified into three categories: hard, intermediate and easy. If there are 7 questions in each category and the test maker is to select two questions from each category, how many different combinations of questions can the test maker put on the test? $$A.\ \ 3\cdot\frac{7!}{5!2!}$$ $$B.\ \ 3\cdot\frac{7!}{5!}$$ $$C.\ \ 3\cdot\frac{7!}{2!}$$ $$D.\ \ 3\cdot21$$ $$E.\ \ 72$$ The OA is A. I'm really confused with this PS question. Experts, any suggestion, please? How can I solve it? Thanks in advance. ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15387 messages Followed by: 1872 members Upvotes: 13060 GMAT Score: 790 LUANDATO wrote: A test maker is to design a probability test from a list of 21 questions. The 21 questions are classified into three categories: hard, intermediate and easy. If there are 7 questions in each category and the test maker is to select two questions from each category, how many different combinations of questions can the test maker put on the test? $$A.\ \ 3\cdot\frac{7!}{5!2!}$$ $$B.\ \ 3\cdot\frac{7!}{5!}$$ $$C.\ \ 3\cdot\frac{7!}{2!}$$ $$D.\ \ 3\cdot21$$ $$E.\ \ 72$$ Hard: From 7 options, the number of ways to choose 2 = 7C2 = (7*6)/(2*1) = 21. Medium: From 7 options, the number of ways to choose 2 = 7C2 = (7*6)/(2*1) = 21. Easy: From 7 options, the number of ways to choose 2 = 7C2 = (7*6)/(2*1) = 21. To combine the options for each category, we multiply: 21*21*21 = 21Â³. None of the answer choices is correct. What is the source? _________________ Mitch Hunt Private Tutor for the GMAT and GRE GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Student Review #1 Student Review #2 Student Review #3 Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. ### GMAT/MBA Expert Legendary Member Joined 14 Jan 2015 Posted: 2666 messages Followed by: 125 members Upvotes: 1153 GMAT Score: 770 LUANDATO wrote: A test maker is to design a probability test from a list of 21 questions. The 21 questions are classified into three categories: hard, intermediate and easy. If there are 7 questions in each category and the test maker is to select two questions from each category, how many different combinations of questions can the test maker put on the test? $$A.\ \ 3\cdot\frac{7!}{5!2!}$$ $$B.\ \ 3\cdot\frac{7!}{5!}$$ $$C.\ \ 3\cdot\frac{7!}{2!}$$ $$D.\ \ 3\cdot21$$ $$E.\ \ 72$$ The OA is A. I'm really confused with this PS question. Experts, any suggestion, please? How can I solve it? Thanks in advance. I'm guessing the question writer intended to write something akin to "The 21 questions are classified into three categories: hard, intermediate and easy. There are 7 questions in each category. The test maker is to select exactly two questions, but both questions must be from the same category. (She can select two questions from the hard category, two questions from the intermediate category, or two questions from the easy category.) How many different ways can she select her pair of questions?" In this case, the answer would have been 7C2 + 7C2 + 7C2, which is equivalent to A. But as Mitch noted, the way the question is currently constructed, none of the answer choices is correct. _________________ Veritas Prep | GMAT Instructor Veritas Prep Reviews Save$100 off any live Veritas Prep GMAT Course

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LUANDATO wrote:
A test maker is to design a probability test from a list of 21 questions. The 21 questions are classified into three categories: hard, intermediate and easy. If there are 7 questions in each category and the test maker is to select two questions from each category, how many different combinations of questions can the test maker put on the test?

$$A.\ \ 3\cdot\frac{7!}{5!2!}$$
$$B.\ \ 3\cdot\frac{7!}{5!}$$
$$C.\ \ 3\cdot\frac{7!}{2!}$$
$$D.\ \ 3\cdot21$$
$$E.\ \ 72$$
Since there are 7 questions in each category and 2 must be selected from each, the number of ways to select the questions is:

7C2 x 7C2 x 7C2

Letâ€™s calculate 7C2:

7C2 = 7!/[2! X (7 - 2) !] = 7!/(2! X 5!) = (7 x 6)/2 = 21

The total number of ways to select the questions is 7C2 x 7C2 x 7C2 = 21^3.

Thus the answer is (7C2)^3 or 21^3. (Note: I donâ€™t see an equivalent answer choice and their choices A and D are actually the same.)

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