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

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

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New post 16 Jul 2017, 11:42
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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


There are two scenarios that fit the question: either one paperback or two is selected.

If one is selected, then the total number of combinations is 2C1*6C3 + 2C2*6C2

2C1*6C3:
2*6!/(3!*3!) = 2*6*5*4/3*2 = 40

2C2*6C2
1*6*5*4*3*2/(2*4*3*2) = 15

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 Jul 2018, 01:09
Could someone provide a group of similar practice problems? Would really appreciate the help on this. Thanks
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Re: There are 8 books in a shelf that consist of 2 paperback [#permalink]

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New post 10 Jul 2018, 02:21
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eishan123 wrote:
Could someone provide a group of similar practice problems? Would really appreciate the help on this. Thanks


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

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New post 11 Jul 2018, 08:51
Thank you! Bunuel
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Re: There are 8 books in a shelf that consist of 2 paperback [#permalink]

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New post 14 Jul 2018, 19:39
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


We can use the equation:

The number of ways in which at least 1 paperback book is selected = The total number of ways to select 4 books – The number of ways in which no paperback books are selected

The number of ways in which no paperback books are selected is equivalent to the number of ways in which all 4 books selected are hardcover. Let’s determine that now. There are 6 hardback books, and 4 must be selected; thus:

6C4 = 6!/(4! x 2!) = (6 x 5)/2! = 30/2 = 15 ways

Now we determine the total number of ways to select the books. There are 8 total books and 4 must be selected, thus:

8C4 = 8!/(4! x 4!) = (8 x 7 x 6 x 5)/4! = 7 x 2 x 5 = 70 ways

Thus, the number of ways to select at least one paperback book = 70 – 15 = 55.

Answer: D
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Re: There are 8 books in a shelf that consist of 2 paperback   [#permalink] 14 Jul 2018, 19:39

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