practice DAT (quant reasoning) question - wrong answer?

[organichemistry]

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This is how it is worded:

Jill has 6 different books. In how many ways can Jill select two different books?

Choices
36
30
18
15
12

I figured 15, but the answer is apparently 30. This is telling me that order, in this case, matters. When reading this question I certainly did not think this way. Any advice on dealing with permutation (and probability) questions like these?

 

[BigJ77]

This is how it is worded:

Jill has 6 different books. In how many ways can Jill select two different books?

Choices
36
30
18
15
12

I figured 15, but the answer is apparently 30. This is telling me that order, in this case, matters. When reading this question I certainly did not think this way. Any advice on dealing with permutation (and probability) questions like these?

I just worked this problem this morning and got 15 and my solution to QR confirmed that the answer was 15. What study source are you using? I think I was using Barron's or Kaplan.
j

 

[Gulch]

This is how it is worded:

Jill has 6 different books. In how many ways can Jill select two different books?

Choices
36
30
18
15
12

I figured 15, but the answer is apparently 30. This is telling me that order, in this case, matters. When reading this question I certainly did not think this way. Any advice on dealing with permutation (and probability) questions like these?

First of all, the answer is 30. We know that since the question is asking for "how many ways..." Therefore, the order in which Jill selects her two book combination matters. You're right, if order didn't matter the answer would be 15, but the facti is that it does. So, there are two ways I can think of going about this problem:

1.) the long way: working out all possible combinations. Let's assign each book a letter: Q,R,S,T,U and V

QR RQ QS SQ QT TQ QU UQ QV VQ RS SR RT TR RU UR RV VR ST TS SU US SV VS TU UT TV VT UV VU (it's saturday, and I had a little extra time 🙂 )

So yeah, Jill could pull out her two books any of these 30 ways.

2.) the short way: permutation style. six different books and two books at a time (four stay on the shelf) which may be in different orders:

6!/ 4! = 6x5 = 30

Hope this helps.

 

[mvs04]

bravo gulch!

I remember this from a Discrete Mathematics class.

 

[dentwannabe]

hah, dun worry about these people talking about Discrete Math and stuff... heres an easier way to think of it.. w/out the factorials and stuff.

say the 6 books are..
A B C D E F

now, two books at a time, lets do it for A..
AB, AC, AD, AE, AF - thats 5 diff. combinations JUST FOR A.
Theres B, C, D, E, F left, each of these will have 5 different combinations. So five different combinations per letter. And there are 6 letters. Therefore, its 5 x 6 = 30

Good Luck.

 

[BigJ77]

hah, dun worry about these people talking about Discrete Math and stuff... heres an easier way to think of it.. w/out the factorials and stuff.

say the 6 books are..
A B C D E F

now, two books at a time, lets do it for A..
AB, AC, AD, AE, AF - thats 5 diff. combinations JUST FOR A.
Theres B, C, D, E, F left, each of these will have 5 different combinations. So five different combinations per letter. And there are 6 letters. Therefore, its 5 x 6 = 30

Good Luck.

Ok, the problem I am working in Barrons (pg 355 #19) is a follows: Kate has six diferent books. Assuming the order doesn't matter, how many ways can Kate select, from these six books, two different books?

The key as mentioned by others is whether order matters or not. In my problem, order doesn't matter so the answer is 15.

My suggestion, get comfortable working them both ways! Thanks for getting me to look back and think also.

j

 

[HBomb]

This is how it is worded:

Jill has 6 different books. In how many ways can Jill select two different books?

Choices
36
30
18
15
12

I figured 15, but the answer is apparently 30. This is telling me that order, in this case, matters. When reading this question I certainly did not think this way. Any advice on dealing with permutation (and probability) questions like these?

The answer you were given (30) is wrong. Just like you said, order doesn't matter, so C(6,2) = 6!/4!2! = 15

If order mattered, then P(6,2)=6!/4! = 30, but if the question was worded just like you stated, then it would seem order doesn't matter...making 30 incorrect.

At any rate, it's good to be critical of answers/solutions. It will help build your understanding of the material.