A test maker is to design a probability test from a list...

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BTGmoderatorLU: Moderator; Posts: 2544; Joined: Sun Oct 15, 2017 1:50 pmFollowed by: 6 members

by BTGmoderatorLU » Fri Feb 02, 2018 6:40 am

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.

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Source: Beat The GMAT — Problem Solving | Fri Feb 02, 2018 6:40 am

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by GMATGuruNY » Fri Feb 02, 2018 8:32 am

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.
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DavidG@VeritasPrep: Legendary Member; Posts: 2663; Joined: Wed Jan 14, 2015 8:25 amFollowed by: 128 members; Location: Boston, MA; Thanked: 1153 times; GMAT Score:770

by DavidG@VeritasPrep » Sun Feb 04, 2018 12:32 pm

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.

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8137; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Mon Feb 05, 2018 9:33 am

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

Let'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 don't see an equivalent answer choice and their choices A and D are actually the same.)

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