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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

by NandishSS » Fri Sep 09, 2016 9:44 am
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

OA:D

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by GMATGuruNY » Fri Sep 09, 2016 9:53 am
NandishSS 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
Combinations with at least 1 paperback = (all possible combinations) - (combinations with no paperbacks).

All possible combinations:
From the 8 books, the number of ways to choose 4 = 8C4 = (8*7*6*5)/(4*3*2*1) = 70.

Combinations with no paperbacks:
From the 6 hardback books, the number of ways to choose 4 = 6C4 = (6*5*4*3)/(4*3*2*1) = 15.

Combinations with at least 1 paperback:
70-15 = 55.

The correct answer is D.
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by Matt@VeritasPrep » Thu Sep 15, 2016 9:25 pm
We could also do

(Exactly One Paperback) + (Exactly Two Paperbacks) =>

(2 choose 1)*(6 choose 3) + (2 choose 2)*(6 choose 2) =>

2*20 + 1*15 =>

55
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by Jeff@TargetTestPrep » Sat Sep 17, 2016 4:33 am
NandishSS 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

OA:D
It must be true that:

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

Therefore,

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

So if we can just determine the number of ways in which no paperback books are selected (that is, all hardback books), and then subtract this value from the total number of ways to select 4 books, we will have our answer.

Let's first determine the number of ways to select all hardback books. There are 6 hardback books, and 4 must be selected; thus:

6C4 = (6 x 5 x 4 x 3)/4! = 3 x 5 = 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 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 ways.

Answer: D

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by hoppycat » Wed May 31, 2017 4:51 am
Matt@VeritasPrep wrote:We could also do

(Exactly One Paperback) + (Exactly Two Paperbacks) =>

(2 choose 1)*(6 choose 3) + (2 choose 2)*(6 choose 2) =>

2*20 + 1*15 =>

55
This looks like the slot approach but it has combinations in it. I though we have to divide by 2! to get rid of double countong. Obviously I'm wrong but I don't get when to divide at the end and when not to
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by DavidG@VeritasPrep » Wed May 31, 2017 6:48 am
hoppycat wrote:
Matt@VeritasPrep wrote:We could also do

(Exactly One Paperback) + (Exactly Two Paperbacks) =>

(2 choose 1)*(6 choose 3) + (2 choose 2)*(6 choose 2) =>

2*20 + 1*15 =>

55
This looks like the slot approach but it has combinations in it. I though we have to divide by 2! to get rid of double countong. Obviously I'm wrong but I don't get when to divide at the end and when not to
The issue is whether the items selected are interchangeable. Notice that in Matt's solution, he found 6 choose 3, by doing (6*5*4)/3!. The 3! accounts for the interchangeability of the items selected. (Choosing Lolita, 1984, and David Copperfield is the same as choosing 1984, David Copperfield and Lolita.) Similarly when he calculated 6 choose 2, he did (6*5)/2! Again, 2! accounts for the fact that the books are interchangeable. So 1) you can use the slot method, and 2) you are dividing the interchangeable components to avoid double-counting.
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by hoppycat » Wed May 31, 2017 12:06 pm
DavidG@VeritasPrep wrote:
hoppycat wrote:
Matt@VeritasPrep wrote:We could also do

(Exactly One Paperback) + (Exactly Two Paperbacks) =>

(2 choose 1)*(6 choose 3) + (2 choose 2)*(6 choose 2) =>

2*20 + 1*15 =>

55
This looks like the slot approach but it has combinations in it. I though we have to divide by 2! to get rid of double countong. Obviously I'm wrong but I don't get when to divide at the end and when not to
The issue is whether the items selected are interchangeable. Notice that in Matt's solution, he found 6 choose 3, by doing (6*5*4)/3!. The 3! accounts for the interchangeability of the items selected. (Choosing Lolita, 1984, and David Copperfield is the same as choosing 1984, David Copperfield and Lolita.) Similarly when he calculated 6 choose 2, he did (6*5)/2! Again, 2! accounts for the fact that the books are interchangeable. So 1) you can use the slot method, and 2) you are dividing the interchangeable components to avoid double-counting.
sweet thanks!
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