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D01-39

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New post 16 Sep 2014, 00:13
2
12
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

69% (02:05) correct 31% (02:28) wrong based on 247 sessions

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New post 16 Sep 2014, 00:13
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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: \(C^5_8+C^4_8*C^1_2=56+140=196\).

Answer: D
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New post 14 Dec 2015, 07:07
"C^5_8 = 56" how is this derived?
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New post 17 Dec 2015, 09:23
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New post 17 Dec 2015, 19:08
Can you explain what the C^5/8 means. How do I get to 56 from this?
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New post 20 Dec 2015, 11:35
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New post 09 Jul 2016, 12:24
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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!
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New post 15 Oct 2016, 11:00
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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
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New post 06 Mar 2017, 21:23
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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
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New post 07 Mar 2017, 02:30
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New post 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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New post 11 Aug 2017, 12:51
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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 = \(140+56 = 196\)
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New post 19 Jun 2018, 22:48
Hi there, i am relatively new to this group and could you please help me with the basics of this topic (may be with some link), i find this really difficult.
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New post 19 Jun 2018, 23:21
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honcho11 wrote:
Hi there, i am relatively new to this group and could you please help me with the basics of this topic (may be with some link), i find this really difficult.


21. Combinatorics/Counting Methods



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New post 31 Jan 2019, 10:16
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Hi Bunuel , Could you please explain to me why is combination used here? Since the books are distinct shouldn't we use permutation i.e. 8*7*6*5*6 (4 small books and 2 large books for the last slot)
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New post 07 Feb 2019, 12:44
This is a great question. I did not think about the possibility of 5 small books. Bookmarked for future practice.
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New post 04 Jun 2019, 12:04
Bunuel wrote:
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

Alternative Solution:

I like to run mine backwards. :-)

(Total Possibilities) \(-\) (Unwanted Outcomes)
Total Possibilities = 10C5 = 252
Unwanted Outcomes = Three Small Books, both Large Books
Three small books: 8C3 = 56
Both large books: 2C2 = 1

= 10C5 \(-\) 8C3(2C2)
= 252 \(-\) 56(1)
= 252 \(-\) 56
= 196 (Wanted Outcomes)
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D01-39   [#permalink] 04 Jun 2019, 12:04
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