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Combination

by mehulv » Mon Jun 21, 2010 8:40 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) (7!/(5!2!))3
B) (7!/(5!))3
C) (7!/(2!))3
D) 3*29
E) 72


OA : A
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by hardik.jadeja » Mon Jun 21, 2010 9:13 am
I feel the answer should be (7!/(5!2!))^3 or 7C2*7C2*7C2, not (7!/(5!2!))3
Last edited by hardik.jadeja on Mon Jun 21, 2010 6:54 pm, edited 2 times in total.
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by outreach » Mon Jun 21, 2010 10:58 am
selectin 2 from 7 is 7C2
NOW 7C2==7!/5!*2!
similalrly 3 level Q means 7C2*7C2*7C2
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by BastiG » Tue Jun 22, 2010 4:10 am
Can someone please explain in more detail why a is correct?
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by newera » Tue Jun 22, 2010 5:57 am
So there are 21 questions total which are grouped into three groups (hard, intermediate, & easy) of seven questions each. You are then told to find the number of ways to pick 2 questions from EACH group (each group has seven questions total).

Because order does NOT matter, this problem is a combination, so you will use "C" NOT "P". So,

Group 1 (Hard questions): 7 questions total. You are choosing 2 --> 7C2 --> 7!/(5!2!)
Group 2 (Intermediate questions): 7 questions total. You are choosing 2 --> 7C2 --> 7!/(5!2!)
Group 3 (Easy questions): 7 questions total. You are choosing 2 --> 7C2 --> 7!/(5!2!)

The next step is then to multiply NOT add. The answer can be written out in simple english as "How many ways can I pick 2 questions from group 1 (hard) AND pick 2 questions from group 2 (intermediate) AND I pick 2 questions from group 3 (easy)?" Multiplication essentially stands for the AND part. Hope that makes sense.

So, the answer is 7C2 * 7C2 *7C2 which is the same as (7C2)^3 which is the same as (7!/(5!2!))^3
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by mehulv » Wed Jun 23, 2010 3:19 am
As the answer is 7C2 the answer should be B as it states that 7!/5!^3 i.e

7*6*5*4*3*2*1/5*4*3*2*1


so that leave us with 7*6 and therefore B


Please tell me what mistake I'm doing


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by newera » Wed Jun 23, 2010 11:18 am
mehulv,

your mistake is that you think 7C2 is 7P2.

Any combination such that rCn = r!/(r-n)!n!. Any permutation such that rPn = r!/(r-n)!

So, 7C2 is correct. The problem is that you wrote it out as a permutation. Make sure to memorize the permutation and combination statements again.

To reiterate, 7C2 = 7!/5!2! = (7*6*5*4*3*2*1)/(5*4*3*2*1*)(2*1)
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