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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

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:

There are 8 books in a shelf that consist of 2 paperback [#permalink]

Show Tags

12 Nov 2007, 01:39

3

This post received KUDOS

14

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

69% (02:29) correct
31% (01:57) wrong based on 882 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?

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

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): \(C^4_6=15\);

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
_________________

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

02 Feb 2014, 18:08

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: There are 8 books in a shelf that consist of 2 paperback [#permalink]

Show Tags

10 Apr 2015, 10:17

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: There are 8 books in a shelf that consist of 2 paperback [#permalink]

Show Tags

16 Apr 2015, 05:30

1

This post received KUDOS

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 8 (so none paperback): \(C^4_6=15\);

{at least one}={total}-{none}=70-15=55.

Answer: D.

Hi Bunuel! There is a typo. Should be "out of 6". Please correct=))
_________________

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 8 (so none paperback): \(C^4_6=15\);

{at least one}={total}-{none}=70-15=55.

Answer: D.

Hi Bunuel! There is a typo. Should be "out of 6". Please correct=))

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

gmatclubot

Re: There are 8 books in a shelf that consist of 2 paperback
[#permalink]
07 May 2015, 05:03

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...