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Difficulty: 605-655 Level,   Combinations,                              
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Re: There are 10 books on a shelf, of which 4 are paperbacks and 6 are har [#permalink]
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Paperbacks - 4, hardbacks - 6

5 books in total and at least 1 from each.

Total combinations for 5 books = (1pb, 4 hb) + (4pb, 1hb) + (3pb, 2hb) + (2pb, 3hb)

1pb, 4hb = 4c1*6c4 = 60
4pb,1hb = 4c4*6c1 = 6
3pb,2hb = 4c3*6c2 = 60
2pb,3hb = 4c2*6c3 = 120

Total combinations of 5 books = 60+6+60+120 = >246

Ans D.
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Re: There are 10 books on a shelf, of which 4 are paperbacks and 6 are har [#permalink]
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JDPB7 wrote:
There are 10 books on a shelf, of which 4 are paperbacks and 6 are hardbacks. How many possible selections of 5 books from the shelf contain at least one paperback and at least one hardback?

A) 75
B) 120
C) 210
D) 246
E) 252



Let us have no restrictions and select 5 books out of available 10 books.

This can be done in 10C5 ways = \(\frac{10!}{5!5!}\) = 252 ways.

Now we need to take out those cases when all books are hardbacks or all books are paperbacks.

Note that when we select all paperbacks, there are only 4 options. So in any case we have to choose one hardback. Condition is satisfied.

So basically we need to subtract only those cases when all hardbacks are chosen.

5 books can be selected out of 6 hardbacks 6C5 ways = \(\frac{6!}{5!1!}\) = 6 ways.

Required answer: 252- 6 = 246 ways

D is the answer.
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Re: There are 10 books on a shelf, of which 4 are paperbacks and 6 are har [#permalink]
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JDPB7 wrote:
There are 10 books on a shelf, of which 4 are paperbacks and 6 are hardbacks. How many possible selections of 5 books from the shelf contain at least one paperback and at least one hardback?

A) 75
B) 120
C) 210
D) 246
E) 252


First let’s determine the total number of ways to select 5 books from 10 books:

10C5 = (10 x 9 x 8 x 7 x 6)/5! = (10 x 9 x 8 x 7 x 6)/(5 x 4 x 3 x 2) = 3 x 2 x 7 x 6 = 252

Since we are selecting 5 books and there are only 4 paperbacks in this 5-book selection, there must be at least 1 hardback. On the other hand, the only way that no paperbacks are in the 5-book selection is when all 5 books are hardback. Since we also want to have at least 1 paperback in the 5-book selection, we need to subtract the number of ways that all 5 hardback books are selected from 252, the total number of ways 5 books can be selected from 10 books.

Number of ways 5 paperback books are selected is:

6C5 = 6

Thus, the total number of ways to select the books with at least 1 paperback and 1 hardback is 252 - 6 = 246 ways.

Answer: D
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Re: There are 10 books on a shelf, of which 4 are paperbacks and 6 are har [#permalink]
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JDPB7 wrote:
There are 10 books on a shelf, of which 4 are paperbacks and 6 are hardbacks. How many possible selections of 5 books from the shelf contain at least one paperback and at least one hardback?

A) 75
B) 120
C) 210
D) 246
E) 252


Ways to choose 1 paperback and 1 handbook

1 P 4H
2P 3H
3P 2H
4P 1H

Basically, there just one way NOT to select at least 1P and 1H- if one selects all 5H

Total ways of selection - ways to select all 5H will give ways to select at least 1P and 1H

10C5- 6C5= 10!/5!5! - 6!/5!= 246

D is the answer
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Re: There are 10 books on a shelf, of which 4 are paperbacks and 6 are har [#permalink]
emi111
you can solve the question by using your method, but it will give you multiple counts for same event. you have to take care of that counts and subtract it. Although your method will give answer, it is lengthy and confusing and strongly not recommended.
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Re: There are 10 books on a shelf, of which 4 are paperbacks and 6 are har [#permalink]
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rakeshpuls123 wrote:
emi111
you can solve the question by using your method, but it will give you multiple counts for same event. you have to take care of that counts and subtract it. Although your method will give answer, it is lengthy and confusing and strongly not recommended.


Please someone elaborate on this.

Why is this logic incorrect - 4 * 6 * 8C3 ?
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There are 10 books on a shelf, of which 4 are paperbacks and 6 are har [#permalink]
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ankur2710 wrote:
There are 10 books on a shelf, of which 4 are paperbacks and 6 are hardbacks. How many possible selections of 5 books from the shelf contain at least one paperback and at least one hardback?

A) 75
B) 120
C) 210
D) 246
E) 252

Please someone elaborate on this.

Why is this logic incorrect - 4 * 6 * 8C3 ?



You will get an answer with duplications in it.

We have 4 paperback and 6 hardback books:
p1, p2, p3, p4, h1, h2, h3, h4, h5, and h6.

One of the cases possible with 4*6*8C3 would be p1 (from 4*6*8C3), h1 (from 4*6*8C3) and p2, p3, p4 (from 4*6*8C3). So, {p1, h1, p2, p3, p4}.

Another possible case would be be p2 (from 4*6*8C3), h1 (from 4*6*8C3) and p1, p3, p4 (from 4*6*8C3). So, {p2, h1, p1, p3, p4}.

But those two cases are the same, which means that with this approach we are getting a lot of cases double-counted.

Hope it's clear.
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Re: There are 10 books on a shelf, of which 4 are paperbacks and 6 are har [#permalink]
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JDPB7 wrote:
There are 10 books on a shelf, of which 4 are paperbacks and 6 are hardbacks. How many possible selections of 5 books from the shelf contain at least one paperback and at least one hardback?

A) 75
B) 120
C) 210
D) 246
E) 252


Key observation: It's impossible to have a selection of 5 books, in which none of the books are hardbacks.
In other words, there will always be at least 1 hardback book in a collection of 5 books, which means we just have to deal with having at least one paperback


Well use to formula: # of ways to obey the restriction = (# of ways to ignore the restriction) - (# of ways to break the restriction)

# of ways to ignore the restriction
In other words, in how many ways can we select 5 books from 10 books?
Since the order in which we select the books does not matter, we can use combinations.
We can select 5 books from 10 books in 10C5 ways
\(10C5 = \frac{(10)(9)(8)(7)(6)}{(5)(4)(3)(2)(1)} = \frac{(9)(8)(7)(6)}{(4)(3)(1)}= \frac{(9)(2)(7)(6)}{(3)(1)}= (3)(2)(7)(6) = 252\)

# of ways to break the restriction
In order to break the restriction, we must have 0 paperbacks in the selection of 5 books,
In other words, we must select 5 books from the 6 hardbacks.
Once again we'll use combinations.
We can select 5 hardbacks from 6 hardbacks in 6C5 ways
\(6C5 = \frac{(6)(5)(4)(3)(2)}{(5)(4)(3)(2)(1)} = 6\)

So, # of ways to obey the restriction \(= 252 - 6 = 246\)

Answer: D
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There are 10 books on a shelf, of which 4 are paperbacks and 6 are har [#permalink]
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Hi GMATters,

Here's my video solution to this problem:



Best,

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Re: There are 10 books on a shelf, of which 4 are paperbacks and 6 are har [#permalink]
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4C1 * 6C1 * 8C3

I understand the complementary route but if we go by this approach, is there a way to remove the duplicates?
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Re: There are 10 books on a shelf, of which 4 are paperbacks and 6 are har [#permalink]
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Vegita wrote:
BrentGMATPrepNow

4C1 * 6C1 * 8C3

I understand the complementary route but if we go by this approach, is there a way to remove the duplicates?


I don't recommend that approach.

Your calculations suggest you are first choosing 1 paperback (in 4C1 ways), and 1 hardback (in 6C1 ways).
This ensures that you have one of each book type (as the question demands)
From there, you choose 3 books from the remaining 8 (thus the 8C3)

At this point, it's VERY difficult (unless I'm missing something) to determine the various ways you have over counted certain outcomes.
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There are 10 books on a shelf, of which 4 are paperbacks and 6 are har [#permalink]
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BrentGMATPrepNow wrote:
Vegita wrote:
BrentGMATPrepNow

4C1 * 6C1 * 8C3

I understand the complementary route but if we go by this approach, is there a way to remove the duplicates?


I don't recommend that approach.

Your calculations suggest you are first choosing 1 paperback (in 4C1 ways), and 1 hardback (in 6C1 ways).
This ensures that you have one of each book type (as the question demands)
From there, you choose 3 books from the remaining 8 (thus the 8C3)

At this point, it's VERY difficult (unless I'm missing something) to determine the various ways you have over counted certain outcomes.


Bunuel BrentGMATPrepNow karishmaB - I didn't quite understand the approach itself though. I've used this multiplicative approach in other problems, and it's usually worked. Why does it break in this case? I used this and ended up answering incorrectly.

Would appreciate your insights!
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Re: There are 10 books on a shelf, of which 4 are paperbacks and 6 are har [#permalink]
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There are 10 books on a shelf, of which 4 are paperbacks and 6 are hardbacks. How many possible selections of 5 books from the shelf contain at least one paperback and at least one hardback?

A) 75
B) 120
C) 210
D) 246
E) 252
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Re: There are 10 books on a shelf, of which 4 are paperbacks and 6 are har [#permalink]
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rahulbiitk wrote:
BrentGMATPrepNow wrote:
Vegita wrote:
BrentGMATPrepNow

4C1 * 6C1 * 8C3

I understand the complementary route but if we go by this approach, is there a way to remove the duplicates?


I don't recommend that approach.

Your calculations suggest you are first choosing 1 paperback (in 4C1 ways), and 1 hardback (in 6C1 ways).
This ensures that you have one of each book type (as the question demands)
From there, you choose 3 books from the remaining 8 (thus the 8C3)

At this point, it's VERY difficult (unless I'm missing something) to determine the various ways you have over counted certain outcomes.


Bunuel BrentGMATPrepNow karishmaB - I didn't quite understand the approach itself though. I've used this multiplicative approach in other problems, and it's usually worked. Why does it break in this case? I used this and ended up answering incorrectly.

Would appreciate your insights!

­That approach doesn't work because the books you choose from for 4C1 and 6C1 are from the same set from which you choose for 8C3. In other words, the other 3 books you do not choose in each case for 4C1 and the other 5 books you do not choose in each case for 6C1 are in the 8C3 set.

So, you end up with multiple ways to choose the same set of five books and thus overcount the number of possible sets of five.
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