Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 02 Apr 2006
Posts: 91

There are 8 books in a shelf that consist of 2 paperback
[#permalink]
Show Tags
Updated on: 02 Feb 2014, 23:48
Question Stats:
71% (01:19) correct 29% (01:40) wrong based on 1260 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
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by pretttyune on 12 Nov 2007, 02:39.
Last edited by Bunuel on 02 Feb 2014, 23:48, edited 3 times in total.
Edited the answer choices and added the OA




Math Expert
Joined: 02 Sep 2009
Posts: 47983

There are 8 books in a shelf that consist of 2 paperback
[#permalink]
Show Tags
08 Feb 2012, 03:20




Director
Joined: 11 Jun 2007
Posts: 889

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
7015 = 55




Senior Manager
Joined: 19 Feb 2007
Posts: 316

No of combinations = p(1) + p(2)
= (2C1 * 6C3) + (2C2 * 6C2)
= (2*6*5*4)/(3*2*1) + (6*5)/2
= 40 + 15
= 55



Manager
Joined: 27 Oct 2008
Posts: 180

Re: (P/S) Combinations
[#permalink]
Show Tags
28 Sep 2009, 04:30
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: 349
Location: Chicago, IL

Re: (P/S) Combinations
[#permalink]
Show Tags
02 Oct 2009, 00:54
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: 105
Location: São Paulo

Re: (P/S) Combinations
[#permalink]
Show Tags
21 Mar 2011, 05:07
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



Retired Moderator
Joined: 16 Nov 2010
Posts: 1458
Location: United States (IN)
Concentration: Strategy, Technology

Re: (P/S) Combinations
[#permalink]
Show Tags
22 Mar 2011, 07:00
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: 13
Location: Taiwan
GMAT 1: 590 Q39 V34 GMAT 2: 680 Q47 V35

Re: There are 8 books in a shelf that consist of 2 paperback
[#permalink]
Show Tags
08 Feb 2012, 01:59
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



Manager
Joined: 10 Jan 2010
Posts: 161
Location: Germany
Concentration: Strategy, General Management
GPA: 3
WE: Consulting (Telecommunications)

Re: There are 8 books in a shelf that consist of 2 paperback
[#permalink]
Show Tags
08 Feb 2012, 05:30
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: 9
Location: India
Concentration: Finance, Marketing
GMAT Date: 11202011
GPA: 3.6
WE: Business Development (Other)

Re: There are 8 books in a shelf that consist of 2 paperback
[#permalink]
Show Tags
08 Feb 2012, 10:00
# 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: 69
Location: United Arab Emirates
Concentration: General Management, Strategy
WE: Supply Chain Management (Energy and Utilities)

Re: There are 8 books in a shelf that consist of 2 paperback
[#permalink]
Show Tags
20 May 2012, 04:36
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: 98

Re: There are 8 books in a shelf that consist of 2 paperback
[#permalink]
Show Tags
25 Jan 2013, 08:20
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
_________________
We appreciate your kudos'



Math Expert
Joined: 02 Sep 2009
Posts: 47983

Re: There are 8 books in a shelf that consist of 2 paperback
[#permalink]
Show Tags
25 Jan 2013, 08:33



Manager
Joined: 18 Oct 2011
Posts: 87
Location: United States
Concentration: Entrepreneurship, Marketing
GMAT Date: 01302013
GPA: 3.3

Re: There are 8 books in a shelf that consist of 2 paperback
[#permalink]
Show Tags
25 Jan 2013, 10:57
# 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)



Intern
Joined: 24 Nov 2014
Posts: 18

Re: There are 8 books in a shelf that consist of 2 paperback
[#permalink]
Show Tags
07 May 2015, 06:03
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



Current Student
Joined: 04 Sep 2012
Posts: 13
Location: United States
Concentration: Finance, Technology
GPA: 3
WE: Securities Sales and Trading (Other)

Re: There are 8 books in a shelf that consist of 2 paperback
[#permalink]
Show Tags
28 Jul 2015, 16:22
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 7015 = 55
CHOOSE D
ON TO THE NEXT



Intern
Joined: 23 Sep 2015
Posts: 39

Re: There are 8 books in a shelf that consist of 2 paperback
[#permalink]
Show Tags
07 Mar 2016, 04:52
a method to solve this problem so there are 2 possibilities either : 1 PB and 3 HB = 2*6*5*4/3! = 40 OR : 2 PB and 2 HB = 2*1*6*5/2!*2! = 15
so we have to add both the possibilities to get the final answer 40+15=55



Manager
Joined: 08 Feb 2016
Posts: 74
Location: India
Concentration: Technology
GPA: 4

There are 8 books in a shelf that consist of 2 paperback
[#permalink]
Show Tags
28 Jul 2016, 06:05
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): \(C^4_6=15\); {at least one}={total}{none}=7015=55. Answer: D. Hi Bunuel, Using PB: Paperbacks ; HB : Hardbacks 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]. So answer is C(8,4) C(6,4) = 7015 = 55



Intern
Joined: 28 Aug 2015
Posts: 19
Location: Norway
WE: Information Technology (Consulting)

Re: There are 8 books in a shelf that consist of 2 paperback
[#permalink]
Show Tags
27 Aug 2016, 03:23
1. Method : 8c4  6c4 = (Total  All four book selected from 6 hard copy)  This will give atleast one paper back selected. 2.Method : 6c2*2c2 + 6c3*2c1 = (PPHH) + (PHHH)
_________________
Thanks Sumit kumar If you like my post Kudos Please




Re: There are 8 books in a shelf that consist of 2 paperback &nbs
[#permalink]
27 Aug 2016, 03:23



Go to page
1 2
Next
[ 26 posts ]



