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There are 8 books in a shelf that consist of 2 paperback [#permalink]
 12 Nov 2007, 02:39
Question Stats:
69% (02:22) correct
30% (02:18) wrong based on 9 sessions
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..
Last edited by Bunuel on 08 Feb 2012, 14:18, edited 2 times in total.
Edited the answer choices and added the OA
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No of combinations = p(1) + p(2)
= (2C1 * 6C3) + (2C2 * 6C2)
= (2*6*5*4)/(3*2*1) + (6*5)/2
= 40 + 15
= 55
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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
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Re: (P/S) Combinations [#permalink]
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
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Re: (P/S) Combinations [#permalink]
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
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Re: (P/S) Combinations [#permalink]
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
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Re: (P/S) Combinations [#permalink]
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
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Re: There are 8 books in a shelf that consist of 2 paperback [#permalink]
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
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Re: There are 8 books in a shelf that consist of 2 paperback [#permalink]
08 Feb 2012, 03:20
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.
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Re: There are 8 books in a shelf that consist of 2 paperback [#permalink]
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
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Re: There are 8 books in a shelf that consist of 2 paperback [#permalink]
08 Feb 2012, 09:16
Here are the ACs:
A. 40
B. 45
C. 50
D. 55
E. 60
FYI.
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Re: There are 8 books in a shelf that consist of 2 paperback [#permalink]
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
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Re: There are 8 books in a shelf that consist of 2 paperback [#permalink]
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
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Re: There are 8 books in a shelf that consist of 2 paperback [#permalink]
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
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Re: There are 8 books in a shelf that consist of 2 paperback [#permalink]
25 Jan 2013, 08:33
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Re: There are 8 books in a shelf that consist of 2 paperback [#permalink]
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)
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Re: There are 8 books in a shelf that consist of 2 paperback
[#permalink]
25 Jan 2013, 10:57
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