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Math Expert V
Joined: 02 Sep 2009
Posts: 58464

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12 00:00

Difficulty:   45% (medium)

Question Stats: 69% (02:05) correct 31% (02:28) wrong based on 247 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 V
Joined: 02 Sep 2009
Posts: 58464

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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: $$C^5_8+C^4_8*C^1_2=56+140=196$$.

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Intern  Joined: 13 Nov 2011
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"C^5_8 = 56" how is this derived?
Math Expert V
Joined: 02 Sep 2009
Posts: 58464

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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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Can you explain what the C^5/8 means. How do I get to 56 from this?
Math Expert V
Joined: 02 Sep 2009
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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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Intern  Joined: 23 Apr 2016
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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!
Intern  Joined: 09 May 2016
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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
Intern  Joined: 08 Mar 2016
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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
Math Expert V
Joined: 02 Sep 2009
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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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Intern  B
Joined: 06 Aug 2017
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I think this is a poor-quality question and I don't agree with the explanation. The answer seems to be wrongly calculated.
Retired Moderator D
Joined: 25 Feb 2013
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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 = $$140+56 = 196$$
Intern  B
Joined: 17 Sep 2013
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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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Math Expert V
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Director  G
Joined: 22 Nov 2018
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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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Manager  S
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This is a great question. I did not think about the possibility of 5 small books. Bookmarked for future practice.
Intern  B
Joined: 14 Mar 2017
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Location: United States (VA)
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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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