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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, 05:23
1
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Question Stats:

61% (01:05) correct 39% (00:52) wrong based on 217 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

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]

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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, 05:38
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1
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, 17:23
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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, 06:25
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1
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, 09:33
1
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, 01:16
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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New post 03 Jan 2019, 10:10
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Replying to a PM :

On first sight, you might fall for the trap that there is only one possibility - 4 small books and 1 large book. However, given information that Romi has space for only one large book (and assuming that one small book can replace one large book by common sense), there are actually two possibilities -

1. 4 small books out of 8 possibilities and 1 large book out of 2 possibilities -

This is in 8C4*2C1 = \(\frac{8!}{(4!*4!)}*2 = \frac{(5*6*7*8)}{(2*3*4)}*2 = 140\) ways

2. 5 small books out of 8 possibilities -

This in 8C5 = \(\frac{8!}{(5!*3!)} = \frac{(6*7*8)}{(2*3)} = 56\) ways.

So a total of (1) or (2)... using the OR-RULE

=140 + 56 = 196 ways.

Hence Option (D) is our bet.

UNSTOPPABLE12, Hope it is clear now.

Let me know if not.
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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 Updated on: 03 Jan 2019, 11:41
Gladiator59 wrote:
Replying to a PM :

On first sight, you might fall for the trap that there is only one possibility - 4 small books and 1 large book. However, given information that Romi has space for only one large book (and assuming that one small book can replace one large book by common sense), there are actually two possibilities -

1. 4 small books out of 8 possibilities and 1 large book out of 2 possibilities -

This is in 8C4*2C1 = \(\frac{8!}{(4!*4!)}*2 = \frac{(5*6*7*8)}{(2*3*4)}*2 = 140\) ways

2. 5 small books out of 8 possibilities -

This in 8C5 = \(\frac{8!}{(5!*3!)} = \frac{(6*7*8)}{(2*3)} = 56\) ways.

So a total of (1) or (2)... using the OR-RULE

=140 + 56 = 196 ways.

Hence Option (D) is our bet.

UNSTOPPABLE12, Hope it is clear now.

Let me know if not.


Perfect explanation, indeed i didnt take into account the two different scenarios, thank you for your time Gladiator59

Originally posted by UNSTOPPABLE12 on 03 Jan 2019, 10:22.
Last edited by UNSTOPPABLE12 on 03 Jan 2019, 11:41, edited 1 time in total.
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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 03 Jan 2019, 10:26
1
TwoThrones wrote:
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

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]


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

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New post 03 Jan 2019, 10:33
1
Just to drill it further -

If you using the slotting technique ( as you pointed out in your PM) you could go ahead by this method -

1. First case (4 small and one large) -

Choosing small books first - 8 ways to choose the first, 7 ways to choose the next, ... 5 ways to choose the 4th.

So (8*7*6*5) / 4! ( we divide by 4! as the order of books is not important)

and choosing one large book in 2 ways.

Hence \(70*2 = 140\) ways.

2. Second case : Now slotting to choose 5 small books -

(8*7*6*5*4) ways to choose the five books and divide by 5! since the order is not important.

Hence in \(\frac{(8*7*6*5*4)}{5!} = \frac{(8*7*6*5*4)}{(2*3*4*5)}= 56\) ways

And then add the two as per OR -rule

Hence a total of 196. By either method.

UNSTOPPABLE12 wrote:
Gladiator59 wrote:
Replying to a PM :

On first sight, you might fall for the trap that there is only one possibility - 4 small books and 1 large book. However, given information that Romi has space for only one large book (and assuming that one small book can replace one large book by common sense), there are actually two possibilities -

1. 4 small books out of 8 possibilities and 1 large book out of 2 possibilities -

This is in 8C4*2C1 = \(\frac{8!}{(4!*4!)}*2 = \frac{(5*6*7*8)}{(2*3*4)}*2 = 140\) ways

2. 5 small books out of 8 possibilities -

This in 8C5 = \(\frac{8!}{(5!*3!)} = \frac{(6*7*8)}{(2*3)} = 56\) ways.

So a total of (1) or (2)... using the OR-RULE

=140 + 56 = 196 ways.

Hence Option (D) is our bet.

UNSTOPPABLE12, Hope it is clear now.

Let me know if not.


Perfect explanation, indeed i didnt take into account the two different scenarios, thank your for your time Gladiator59

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Regards,
Gladi



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Re: Romi has a collection of 10 distinct books (8 small and 2 large). In &nbs [#permalink] 03 Jan 2019, 10:33
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