Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
In case you didn’t notice, we recently held the 1st ever GMAT game show and it was awesome! See who won a full GMAT course, and register to the next one.
There are 8 books in a shelf that consist of 2 paperback
[#permalink]
Show Tags
Updated on: 02 Feb 2014, 22:48
10
44
00:00
A
B
C
D
E
Difficulty:
35% (medium)
Question Stats:
71% (01:57) correct 29% (02:20) wrong based on 1355 sessions
HideShow timer Statistics
There are 8 books in a shelf that consist of 2 paperback books and 6 hardback books. If 4 books are selected at random without replacement, how many different combinations are there that at least one paperback book is selected?
There are 8 books in a shelf that consist of 2 paperback
[#permalink]
Show Tags
08 Feb 2012, 02:20
19
17
pretttyune wrote:
There are 8 books in a shelf that consist of 2 paperback books and 6 hardback books. If 4 books are selected at random without replacement, how many different combinations are there that at least one paperback book is selected?
A. 40 B. 45 C. 50 D. 55 E. 60
Thanks a lot for your help in advance..
It's almost always better to solve "at least one" combinations/probability questions with {at least one}={total}-{none}.
Total # of selections of 4 books out of 8: \(C^4_8=70\); Total # of selections of 4 hardback books out of 6 (so none paperback): \(C^4_6=15\);
I was taking the approach of first finding out the number of ways of selecting 4 books out of 8, then finding the number of possibilities where no paperbacks were chosen, and subtracting the two.
If anyone can help with this approach to put down what the actual terms should be, thatd be very helpful to me.
There are 8 books in a shelf that consist of 2 paperback books and 6 hardback books. If 4 books are selected at random without replacement, how many different combinations are there that at least one paperback book is selected ?
Soln: No. of combinations which have atleast one paperback is = Total no. of combinations - Combinations in which there is no paperback = 8C4 - 6C4 = 55 ways
Other way of doing this is finding out directly
No. of combinations which have atleast one paperback is = Total no. of combinations that have 2 paperback + total no. of combinations that have 1 paperback = 2C2 * 6C2 + 2C1 * 6C3 = 15 + 40 = 55 ways
Re: There are 8 books in a shelf that consist of 2 paperback
[#permalink]
Show Tags
08 Feb 2012, 09:00
# of selections of books with no condition = 8C4 = 70 # of selections of books with no paperback book = 6C4 = 15 # of selections of books with at least one paperback book = 70 -15 = 55
Re: There are 8 books in a shelf that consist of 2 paperback
[#permalink]
Show Tags
25 Jan 2013, 07:20
Hmm....so when I did this I made an error that might have carried over from using combinatorics from probability. I found the total as 8C4 and to find the amount to subtract from it I had (6C4)(2C0). Why don't we multiply by the books not chosen and what's would it mean if we did?
I'm trying to understand a fundamental flaw I made here
_________________
Re: There are 8 books in a shelf that consist of 2 paperback
[#permalink]
Show Tags
25 Jan 2013, 07:33
1
manimgoindowndown wrote:
Hmm....so when I did this I made an error that might have carried over from using combinatorics from probability. I found the total as 8C4 and to find the amount to subtract from it I had (6C4)(2C0). Why don't we multiply by the books not chosen and what's would it mean if we did?
I'm trying to understand a fundamental flaw I made here
You'd get the same result because 2C0=2!/(2!*0!)=2/(2*1)=1: there is one way to choose 0 books from 2 books.
Re: There are 8 books in a shelf that consist of 2 paperback
[#permalink]
Show Tags
07 May 2015, 05:03
When I first read the question I understood that we need to find the number of ways to select 4 books with at least 1 of them being a paperback, instead of calculating the number of possible selections, were PHHH = HPHH
Would the answer to this other question be correct, as this might show up on the exam? 2 Paperbacks: 2*1*6*5 = 60 1 Paperback : 1*6*5*4 = 120 = 180
Re: There are 8 books in a shelf that consist of 2 paperback
[#permalink]
Show Tags
28 Jul 2015, 15:22
First thing to notice when you look at this question: we're dealing with permuations/combinations. Which to use? Ranking/order doesn't matter here - which means we use combinations.
Next thing to notice: at least one - i don't know anyone who likes these questions! Strategy should be TOTAL - NONE = AT LEAST 1 (our desired solution)
TOTAL = 8c4 (8 books 4 being selected) = 70 NONE = 6c4 (Only 6 books because we aren't including paperback, 4 being selected) = 15 70-15 = 55
There are 8 books in a shelf that consist of 2 paperback
[#permalink]
Show Tags
28 Jul 2016, 05:05
Bunuel wrote:
pretttyune wrote:
There are 8 books in a shelf that consist of 2 paperback books and 6 hardback books. If 4 books are selected at random without replacement, how many different combinations are there that at least one paperback book is selected?
A. 40 B. 45 C. 50 D. 55 E. 60
Thanks a lot for your help in advance..
It's almost always better to solve "at least one" combinations/probability questions with {at least one}={total}-{none}.
Total # of selections of 4 books out of 8: \(C^4_8=70\); Total # of selections of 4 hardback books out of 6 (so none paperback): \(C^4_6=15\);
If the question were - "How many possible selections of 4 books include both the PB" then is why is my following solution wrong ?
Total possibility .: C(8,4) Possibility of 2 PB = Total possibility - possibility of no PB => C(8,4)- C(6,4) [C(6,4) -> removing 2 PB from the lot will leave only 6 books. So selecting 4 books from 6].
close
71% (01:57) correct 
