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

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

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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, 23:48, edited 3 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]

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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]

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New post 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]

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New post 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

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Re: (P/S) Combinations [#permalink]

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New post 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]

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

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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}=70-15=55.

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

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New post 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]

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New post 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]

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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]

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New post 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]

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

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New post 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]

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New post 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

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

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New post 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
70-15 = 55

CHOOSE D

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

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New post 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

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

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New post 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}=70-15=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) = 70-15 = 55

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

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

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