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Romi has a collection of 10 distinct books (8 small and 2 large). In

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Romi has a collection of 10 distinct books (8 small and 2 large). In [#permalink]

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New post 24 Feb 2011, 06:23
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Question Stats:

64% (00:56) correct 36% (00:58) wrong based on 100 sessions

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Romi has a collection of 10 distinct books (8 small and 2 large). In how many ways can he select 5 books to take with him on a trip if he has room for only 1 large book?

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

[Reveal] Spoiler:
GMAT Club Diagnostic Test PS Question #40

I have the answer listed with the explanation below, but I still do not understand how they arrived with that.

[spoiler=]The number of ways 5 books can be chosen (with max. 1 large book) equals (total number of ways 5 books can be chosen) - (2 large + 3 small books will be chosen).

Total number of ways 5 books can be chosen \(= C_{10}^5 = \frac{10*9*8*7*6*5!}{5!5!} = 252\) .

Number of possible ways that 3 small and 2 large books will be chosen \(= 1*C_8^3 = 1 * 56 = 56\) .

Desired number of ways that 5 books can be taken \(= 252 - 56 = 196\) .


If someone can explain this either with the slot method or any other method from the Manhattan Guide Word Translations, I would greatly appreciate it. Thank you very much![/spoiler]
[Reveal] Spoiler: OA

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Re: Romi has a collection of 10 distinct books (8 small and 2 large). In [#permalink]

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New post 24 Feb 2011, 06:38
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Sol:

In how many ways can Romi select 5 books:

5 Small books out of 8 AND 0 large book out of 2 books
OR
4 Small book out of 8 AND 1 large book out of 2 books

We know AND represents multiplication and OR addition;

Thus; to convert the above statement into combinations:

\(C^8_5*C^2_0 + C^8_4*C^2_1\)

\(\frac{8!}{5!3!}*\frac{2!}{2!0!} + \frac{8!}{4!4!}*\frac{2!}{1!1!}\)

\(56 + 140 = 196\)

Ans: "D"

*************************

Sorry, I have no idea about Slot Method!!!
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Re: Romi has a collection of 10 distinct books (8 small and 2 large). In [#permalink]

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New post 24 Feb 2011, 18:23
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If he is to select 4 small books and 1 large book, then the number of arrangements is: 8! / 4!4! * 2!/1!1!, where the first term is the number of ways in which he can pick the 4 small books and the second term is the number of ways in which he can pick the large books. This results 140. However, he can also pick 5 small books. The number of arrangements in which he can pick the 5 books will be 8!/5!3!, which results in 8*7 = 56.

Now, you have to sum both possibilities, which yields 196.

Hope it helped.

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Re: Romi has a collection of 10 distinct books (8 small and 2 large). In [#permalink]

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New post 25 Feb 2011, 07:25
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Here is how you understand the given solution.
We just want to make sure that out of the 2 larger books , we don't select both of them (we can select 0 or 1 but not both).
So the number of ways = select any 5 books - select both the larger books and any 3 smaller books.

Hope that explains your question.

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Re: Romi has a collection of 10 distinct books (8 small and 2 large). In [#permalink]

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New post 25 Feb 2011, 10:33
Ladies and gentlemen! I highly appreciate your help. It seems using the Manhattan GMAT Word Translations strategy was too tricky for me. I used the nCr approach instead of the slot method (I had to do a little research because I hadn't done this since high school) and found out how much easier the question was. Thank you very much everyone!
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Re: Romi has a collection of 10 distinct books (8 small and 2 large). In [#permalink]

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New post 26 Feb 2011, 02:16
Yeah first select 5 books i.e. 10C5 and then subtract the remainder i.e. All big + 3 small
10C5 - 8C3 = 196

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Re: Romi has a collection of 10 distinct books (8 small and 2 large). In [#permalink]

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Re: Romi has a collection of 10 distinct books (8 small and 2 large). In   [#permalink] 09 Oct 2017, 10:38
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