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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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06 Nov 2015, 17:43

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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?

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

the best way would be..

1) choosing ways where all 5 are of one kind and subtracting from all ways... total ways- \(10C5\) all one of a kind-\(6C5\) total selections=\(\frac{10!}{5!5!}-\frac{6!}{5!1!}=246\) 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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09 Nov 2015, 16:26

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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

You can calculate directly possible selections of 5 books from the shelf that contain at least one paperback and at least one hardback directly as shyind did or indirectly as chetan2u did.

I will elaborate further on chetan2u's method. To calculate the possible selections of 5 books from the shelf that contain at least one paperback and at least one hardback, subtract the complement of at least one paperback and at least one hardback from total number of 5 book combinations. To find the complement event, use DeMorgan's rule.

DeMorgan's Rule states that ~(P and Q) = ~P or ~Q ~ symbol means Not

So if we let P = at least one paperback then ~P = No paperbacks. So if we let Q = at least one hardback then ~Q = No hardbacks.

No hardbacks would mean 5 paperbacks, but this is not possible since there are only 4 paperbacks. ~Q = 0 ways No paperbacks would mean 5 hardbacks, which can be chosen in \(C^6_5\) in 6 ways. ~P = 6 ways ~P or ~Q = ~P + ~Q = 6 + 0 = 6 ways

Total number of 5 book combinations from 10 books is \(C^{10}_5\) = \(\frac{(10*9*8*7*6)}{(5*4*3*2*1)}\) = 252 ways

(Total number of 5 book combinations) - (No paperbacks or No hardbacks) = 252 - 6 = 246.

For a total of 244, which is clearly wrong. Where did I go wrong? the total is 264, which is also wrong

2pb, 3 hb: 4! x 6! = 120 2!2! 3!3!

4pb, 1hb 4 x 6 = 24

Thanks in advance

Hi, the mistake is in the coloured portion above.. choosing 4 paperback and 1 hardback.. 4C4*6C1= 1*6=6.. you have taken 4C4 as 4.. so the answer now is= 60+60+120+6=246.. hope it helps
_________________

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

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13 Jun 2016, 14:08

chetan2u wrote:

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

the best way would be..

1) choosing ways where all 5 are of one kind and subtracting from all ways... total ways- 10C5 all one of a kind-6C5 total selections=10!/5!5!-6!/5!1!=246 D

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

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13 Jun 2016, 16:30

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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
_________________

I welcome critical analysis of my post!! That will help me reach 700+

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

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13 Jun 2016, 20:52

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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.

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

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03 Jan 2017, 01:39

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.

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

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06 Jan 2017, 10:50

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.

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

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10 Mar 2017, 20:38

ankur2710 wrote:

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 ?

I have the same question. Can Bunuel please help? Thanks!

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 ?

I have the same question. Can Bunuel please help? Thanks!

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.

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

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11 Mar 2017, 07:28

Bunuel wrote:

golden999 wrote:

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 ?

I have the same question. Can Bunuel please help? Thanks!

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.

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

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23 Mar 2017, 03:51

My approach was to find total # of combination to choose 5 books out of 10 -> 5C10=252 then to deduct combinations of choosing all hardcover 5C6=6 and of papercover 5C4 =0. Hence 252-6=246, answer is D

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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14 May 2017, 00:57

adiagr wrote:

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.

yes thats the easiest way imo

Nonetheless I couldn't remember how to count the numbers of "all hardbacks"

maybe a hint in the stem for this strategy is the wording "at least"
_________________

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

1. Ordering is not important and there is no repetition, so this is a nCr problem 2. The constraint is given which is selecting at least one paperback and atleast one hardback 3. The opposite of the constraint is easier to find which is no hardback or no paperback . But 1 hardback has to be there out of 5 books selected, since there are only 4 paperbacks. so the opposite of the constraint effectively boils down to no paperback 4. Total number of combinations is 10C5 5. Number of combinations without constraints is selecting 5 hardbacks out of six hardbacks which is 6C5 6. Number of combinations with constraints is (4)-(5)= 246
_________________

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

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12 Nov 2017, 05:04

ramonguib wrote:

chetan2u wrote:

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

the best way would be..

1) choosing ways where all 5 are of one kind and subtracting from all ways... total ways- 10C5 all one of a kind-6C5 total selections=10!/5!5!-6!/5!1!=246 D

Hello, chetan2u, If there were 5 paperbacks and 5 hardbacks, would this be correct: Total ways = 10C5 = 252 All of a kind = (paperbacks) 5C5 = 1 All of a kind = (hardbacks) 5CC = 1 total selecions = 252 - 1 - 1 = 250 ? Thanks!

I would be greatful, if math experts could confirm whether it would be right solution for 5 paperbacks and 5 hardbacks....