There are 8 books in a shelf that consist of 2 paperback : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 23 Jan 2017, 21:31

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 02 Apr 2006
Posts: 93
Followers: 1

Kudos [?]: 36 [4] , given: 0

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

### Show Tags

12 Nov 2007, 01:39
4
KUDOS
13
This post was
BOOKMARKED
00:00

Difficulty:

35% (medium)

Question Stats:

69% (02:28) correct 31% (01:56) wrong based on 907 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?

A. 40
B. 45
C. 50
D. 55
E. 60
[Reveal] Spoiler: OA

Last edited by Bunuel on 02 Feb 2014, 22:48, edited 3 times in total.
Senior Manager
Joined: 19 Feb 2007
Posts: 325
Followers: 1

Kudos [?]: 46 [1] , given: 0

### Show Tags

12 Nov 2007, 02:12
1
KUDOS
2
This post was
BOOKMARKED
No of combinations = p(1) + p(2)

= (2C1 * 6C3) + (2C2 * 6C2)

= (2*6*5*4)/(3*2*1) + (6*5)/2

= 40 + 15

= 55
Director
Joined: 11 Jun 2007
Posts: 931
Followers: 1

Kudos [?]: 175 [5] , given: 0

### Show Tags

12 Nov 2007, 19:27
5
KUDOS
3
This post was
BOOKMARKED
pmenon wrote:
Is the answer above right ?

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.

i would go about it this way:

total - all hardback
total = 8C4
hardback = 6C4

8C4 - 6C4
70-15 = 55
Manager
Joined: 27 Oct 2008
Posts: 185
Followers: 2

Kudos [?]: 144 [3] , given: 3

### Show Tags

28 Sep 2009, 03:30
3
KUDOS
3
This post was
BOOKMARKED
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
Senior Manager
Joined: 03 Nov 2005
Posts: 396
Location: Chicago, IL
Followers: 3

Kudos [?]: 50 [0], given: 17

### Show Tags

01 Oct 2009, 23:54
Solution:

Total possible combinations of books = C(8,4)

Total possible combinations without paperback books C(6,4)

Total possible combinations with at least 1 paperback book: C(8,4)-C(6,4)=55
_________________

Hard work is the main determinant of success

Manager
Joined: 01 May 2008
Posts: 114
Location: São Paulo
Followers: 1

Kudos [?]: 6 [0], given: 0

### Show Tags

21 Mar 2011, 04:07
Another approach:

Combinations with the first book:
6C3 = 20

Combinations with the second book:
6C3 = 20

Combinations with both books:
6C2 = 15

Total = 20+20+15 = 55

Regards
SVP
Joined: 16 Nov 2010
Posts: 1672
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 33

Kudos [?]: 514 [0], given: 36

### Show Tags

22 Mar 2011, 06:00
PHHH + PPHH

= 2C1 * 6C3 + 2C2 * 6C2

= 2 * 6!/3!3! + 1 * 6!/4!2!

= 2* 6*5*4/3! + 6 * 5/2

= 40 + 15 = 55
_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

GMAT Club Premium Membership - big benefits and savings

Intern
Joined: 17 Oct 2011
Posts: 27
Location: Taiwan
GMAT 1: 590 Q39 V34
GMAT 2: 680 Q47 V35
Followers: 0

Kudos [?]: 49 [0], given: 10

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

### Show Tags

08 Feb 2012, 00:59
I came up with a different (more complicated?) approach.

FIRST STEP:
How many possible selections of 4 books from this self including NO paperback?
6/8 * 5/7 * 4/6 * 3/5 = 3/14

SECOND STEP:
Subtract 3/14 from 1 (or 14/14) => 14/14 - 3/14 = 11/14

THIRD STEP:
The only answer choice with 11 as a factor is 55.

Ans. C
Math Expert
Joined: 02 Sep 2009
Posts: 36618
Followers: 7102

Kudos [?]: 93591 [9] , given: 10578

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

### Show Tags

08 Feb 2012, 02:20
9
KUDOS
Expert's post
10
This post was
BOOKMARKED
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

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): $$C^4_6=15$$;

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

_________________
Manager
Joined: 10 Jan 2010
Posts: 192
Location: Germany
Concentration: Strategy, General Management
Schools: IE '15 (M)
GMAT 1: Q V
GPA: 3
WE: Consulting (Telecommunications)
Followers: 2

Kudos [?]: 28 [0], given: 7

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

### Show Tags

08 Feb 2012, 04:30
Do i miss the answer choices?

total combinations: $$C^4_8 = 70$$
No Paperback: $$C^4_6 = 15$$

Total - No Paperback = At least 1 Paperback
70 - 15 = 55 combinations
Intern
Joined: 29 Jan 2011
Posts: 17
Location: India
Concentration: Finance, Marketing
Schools: HKUST, ISB
GMAT Date: 11-20-2011
GPA: 3.6
Followers: 0

Kudos [?]: 11 [0], given: 7

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
Manager
Status: Sky is the limit
Affiliations: CIPS
Joined: 01 Apr 2012
Posts: 56
Location: United Arab Emirates
Concentration: General Management, Strategy
GMAT 1: 650 Q49 V31
GMAT 2: 720 Q50 V38
WE: Supply Chain Management (Energy and Utilities)
Followers: 1

Kudos [?]: 28 [3] , given: 10

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

### Show Tags

20 May 2012, 03:36
3
KUDOS
The selection should at least contain one PAPERBACK book. There are a total of 2 PB and 6 HB books are available.

The combinations that at least one PB book will come out are: PHHH & PPHH

1. PHHH = 2C1 * 6C3 = 2 * (5*4) = 40
2. PPHH = 2C2 * 6C2 = 1 * ((6*5) / 2) = 15

In total 40+15 = 55 Ways
Manager
Joined: 07 Feb 2011
Posts: 92
Followers: 0

Kudos [?]: 59 [0], given: 44

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
_________________

Math Expert
Joined: 02 Sep 2009
Posts: 36618
Followers: 7102

Kudos [?]: 93591 [1] , given: 10578

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

### Show Tags

25 Jan 2013, 07:33
1
KUDOS
Expert's post
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.

Hope it's clear.
_________________
Manager
Joined: 18 Oct 2011
Posts: 90
Location: United States
Concentration: Entrepreneurship, Marketing
GMAT Date: 01-30-2013
GPA: 3.3
Followers: 2

Kudos [?]: 69 [0], given: 0

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

### Show Tags

25 Jan 2013, 09:57
# of ways of choosing 4 books from 8 total books = 8C4 = 70
# of ways of only choosing hardback books = 6C4 = 15

Therefore, # of ways of choosing 4 books with at least 1 paperback book = 8C4 - 6C4 = 55 (D)
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13532
Followers: 577

Kudos [?]: 163 [0], given: 0

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.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13532
Followers: 577

Kudos [?]: 163 [0], given: 0

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.
_________________
Senior Manager
Joined: 02 Dec 2014
Posts: 388
Location: Russian Federation
Concentration: General Management, Economics
GMAT 1: 640 Q44 V33
WE: Sales (Telecommunications)
Followers: 0

Kudos [?]: 79 [1] , given: 349

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

### Show Tags

16 Apr 2015, 05:30
1
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

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.

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

"Are you gangsters?" - "No we are Russians!"

Math Expert
Joined: 02 Sep 2009
Posts: 36618
Followers: 7102

Kudos [?]: 93591 [0], given: 10578

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

### Show Tags

16 Apr 2015, 05:34
Konstantin1983 wrote:
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

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.

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

Typo edited. Thank you.
_________________
Intern
Joined: 24 Nov 2014
Posts: 18
Followers: 0

Kudos [?]: 2 [0], given: 0

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] 07 May 2015, 05:03

Go to page    1   2    Next  [ 24 posts ]

Similar topics Replies Last post
Similar
Topics:
34 There are 10 books on a shelf, of which 4 are paperbacks and 6 are har 12 06 Nov 2015, 17:43
5 There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 13 13 Jul 2015, 12:15
2 On a shelf there are 6 hardback books and 2 paperback book. 2 30 Jun 2012, 06:18
2 Of the books standing in a row on a shelf, an atlas is the 3 14 Jan 2012, 07:47
10 There are 8 books in a shelf that consist of 2 paperback books and 6 9 05 Jul 2007, 19:43
Display posts from previous: Sort by