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# D01-39

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Math Expert
Joined: 02 Sep 2009
Posts: 46297

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16 Sep 2014, 00:13
2
9
00:00

Difficulty:

55% (hard)

Question Stats:

66% (01:20) correct 34% (01:59) wrong based on 179 sessions

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Romi has a collection of 10 distinct books out of which 8 are small and 2 are large. In how many ways can he select 5 books to take with him on a trip if he can take at most 1 large book?

A. 56
B. 126
C. 152
D. 196
E. 252

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Math Expert
Joined: 02 Sep 2009
Posts: 46297

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16 Sep 2014, 00:13
2
2
Official Solution:

Romi has a collection of 10 distinct books out of which 8 are small and 2 are large. In how many ways can he select 5 books to take with him on a trip if he can take at most 1 large book?

A. 56
B. 126
C. 152
D. 196
E. 252

Since Romi can take at most 1 large book then he can take with him either 5 small books or 4 small books and 1 large book: $$C^5_8+C^4_8*C^1_2=56+140=196$$.

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Intern
Joined: 11 Jan 2015
Posts: 3
GPA: 3.3

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13 Jan 2015, 11:12
Could you pls explain this math in other way except

Since Romi can take at most 1 large book then he can take with him either 5 small books or 4 small books and 1 large book: C^5_8+C^4_8*C^1_2=56+140=196.
Intern
Joined: 13 Nov 2011
Posts: 18

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14 Dec 2015, 07:07
"C^5_8 = 56" how is this derived?
Math Expert
Joined: 02 Sep 2009
Posts: 46297

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17 Dec 2015, 09:23
LostinNY wrote:
"C^5_8 = 56" how is this derived?

Choosing 5 small books out of 8. So, Romi will have 5 small books and 0 large books.
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Joined: 08 Nov 2015
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17 Dec 2015, 19:08
Can you explain what the C^5/8 means. How do I get to 56 from this?
Math Expert
Joined: 02 Sep 2009
Posts: 46297

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20 Dec 2015, 11:35
dau245 wrote:
Can you explain what the C^5/8 means. How do I get to 56 from this?

C stands for combinations. Check for more here: math-combinatorics-87345.html
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Joined: 31 Oct 2015
Posts: 9
Concentration: Real Estate, Social Entrepreneurship
GMAT 1: 720 Q48 V42
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06 May 2016, 16:21
Based on the combinatorics post Cn_k = n!/(k!(n-k)?)

Doesn't the solution provided switch the n and the k? Just learning this so maybe I am mistaken. Thanks!
Intern
Joined: 23 Apr 2016
Posts: 14

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09 Jul 2016, 12:24
2
See example in the ebook from Magoosh team under Combinations: http://magoosh.resources.s3.amazonaws.c ... _eBook.pdf

C 5 over 8 = 8!/5!(8-5)! = 8*7*6*5*4*3*2*1/(5*4*3*2*1)*(3*2*1)
Simplify and you get 8*7*6/3*2*1
Simplify more and you get 4*7*2 = 8*7 = 56

You do the same with the other part of the equation.
So this for the first part. Hope this helps everyone!
Intern
Joined: 09 May 2016
Posts: 2

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15 Oct 2016, 11:00
1
Alternative solution
Since Romi can take at most 1 large book then he can take with him : All possible ways - (2 Large books and 3 books from 5 possible) =(5 small books from 10) - (3 small books from 8)=C510-C38=196
Current Student
Joined: 23 Jan 2016
Posts: 12
GMAT 1: 640 Q36 V40
GMAT 2: 650 Q41 V38
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05 Nov 2016, 22:42
Nawfal wrote:
Alternative solution
Since Romi can take at most 1 large book then he can take with him : All possible ways - (2 Large books and 3 books from 5 possible) =(5 small books from 10) - (3 small books from 8)=C510-C38=196

Thanks Nawfal, that was the way I went about solving as well!
Intern
Joined: 08 Mar 2016
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06 Mar 2017, 21:23
1
I have a question, for the second combination C^4_8, I get 70 not 140.
C^4_8 = 8*7*6*5 / 4*3*2*1 = 2*7*5 = 70
Math Expert
Joined: 02 Sep 2009
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07 Mar 2017, 02:30
jdlopez0224 wrote:
I have a question, for the second combination C^4_8, I get 70 not 140.
C^4_8 = 8*7*6*5 / 4*3*2*1 = 2*7*5 = 70

Yes, 8C4 = 70, but there we have 8C4 * 2C1 = 70*2 = 140.
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Joined: 07 Feb 2016
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GMAT 1: 650 Q47 V34
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01 May 2017, 23:07
Although it is more slowly (as it aims for probability and not for combinations), I want to add a different approach.

$$P(at least one large book)=1-P(two large books)$$

$$P(two large books)=\frac{2}{10}*\frac{1}{9}*C^2_5=\frac{1}{45}*10=\frac{2}{9}$$

$$P(at least one large book)=1-P(two large books)=\frac{7}{9}$$

$$Combinations=P(at least one large book)*C^5_10=\frac{7}{9}*252=196$$
Intern
Joined: 06 Aug 2017
Posts: 1

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11 Aug 2017, 12:17
I think this is a poor-quality question and I don't agree with the explanation. The answer seems to be wrongly calculated.
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Joined: 25 Feb 2013
Posts: 1143
Location: India
GPA: 3.82

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11 Aug 2017, 12:51
Victor__ wrote:
I think this is a poor-quality question and I don't agree with the explanation. The answer seems to be wrongly calculated.

Hi Victor__

it would be great if you could explain what exactly is wrong with the official solution?

As per the question there are two possibilities -

A. take $$1$$ large book $$AND$$ $$4$$ small books. Mathematically this can be represented as -
$$1$$ large book can be selected from $$2$$ large books in $$2_C_1$$ and remaining $$4$$ small books can be selected from $$8$$ small books in $$8_C_4$$
therefore total number of ways = $$2_C_1$$ $$*$$ $$8_C_4 = 140$$

$$OR$$

B. take $$5$$ small books and no large books. Mathematically it can be represented as -
$$5$$ small books can be selected from $$8$$ books in $$8_C_5 = 56$$

Therefore total number of ways = $$140+56 = 196$$
Intern
Joined: 22 Jan 2018
Posts: 2

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04 Feb 2018, 14:52
Since we can choose 1 Large book or 0 Large book then:

L-S-S-S-S/4! + S-S-S-S-S/5! (we divide for n! each sub set, n=number of positions)

(2/1!) * (8*7*6*5/4!) + (8*7*6*5*4/5!) = 196
Intern
Joined: 09 Sep 2015
Posts: 25

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01 Apr 2018, 07:04
Hi Guys

What if the question asks for "at least 1 large book" then in how many ways Romi can select 5 books ?

Any Help is appreciated

Thanks
PS Forum Moderator
Joined: 25 Feb 2013
Posts: 1143
Location: India
GPA: 3.82

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01 Apr 2018, 09:06
Hi Guys

What if the question asks for "at least 1 large book" then in how many ways Romi can select 5 books ?

Any Help is appreciated

Thanks

At least 1 large book will mean that there will be two scenarios -

Scenario 1: 1 Large book and 4 Small Books =$$2_C_1 * 8_C_4$$ OR

Scenario 2: 2 Large books and 3 Small Books =$$2_C_2*8_C_3$$

So the final answer will be the sum of these two scenarios
Intern
Joined: 09 Sep 2015
Posts: 25

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01 Apr 2018, 21:07
niks18 wrote:
Hi Guys

What if the question asks for "at least 1 large book" then in how many ways Romi can select 5 books ?

Any Help is appreciated

Thanks

At least 1 large book will mean that there will be two scenarios -

Scenario 1: 1 Large book and 4 Small Books =$$2_C_1 * 8_C_4$$ OR

Scenario 2: 2 Large books and 3 Small Books =$$2_C_2*8_C_3$$

So the final answer will be the sum of these two scenarios

Thanks niks18
Re: D01-39   [#permalink] 01 Apr 2018, 21:07

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# D01-39

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