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# A book publisher received 4 horror stories, 5 romantic

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05 Feb 2004, 08:53
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A book publisher received 4 horror stories, 5 romantic stories and 3 comedy stories. He wants to put together a book, contemporary stories, that should have at least one romantic story and one comedy story. How many different combinations are possible?
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05 Feb 2004, 09:57
A book publisher received 4 horror stories, 5 romantic stories and 3 comedy stories. He wants to put together a book, contemporary stories, that should have at least one romantic story and one comedy story.

So, the book must come with a romantic story and a comedy story-- we can add up to:
4 horror stories-- 1 way to choose zero, 4 ways to choose one, 6 ways to choose two, 4 ways to choose three, 1 way to choose four. 16 configurations
4 romantic stories-- 1 way to choose zero, 4 ways to choose one, 6 ways to choose two, 4 ways to choose three, 1 way to choose four. 16 configurations
3 comedy stories-- 1 way to choose zero, 3 ways to choose one, 3 ways to choose two, 1 way to choose three. 8 configurations.

16*16*8

2048.

I know there's a substantially easier way to solve this, but it escapes me right now.
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05 Feb 2004, 10:02
(2^5 - 1) * (2^3 - 1) * (2^4) = 3472
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05 Feb 2004, 10:10
gmatblast- you are right. I see my mistake.
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05 Feb 2004, 11:28
gmatblast

how did you come up with this formula. can you please exaplain in more details.

My line of thinking was

At lease one romantic and one comedy can be arrenged in 5*3 = 15
and then the 4 horror could be 15*5 ways or books
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05 Feb 2004, 11:56
pakoo wrote:
gmatblast

how did you come up with this formula. can you please exaplain in more details.

My line of thinking was

At lease one romantic and one comedy can be arrenged in 5*3 = 15
and then the 4 horror could be 15*5 ways or books

Pakoo,

What you have done is I guess, number of combinations IF ALL THE STORIES ARE TO BE USED. And it does not reflect the given retstriction properly.

We do not need to use all the stories. The only constraint that is applicable here is we must have 1R + 1C in whatever combination we choose.

That being said,

The reasoning I applied is that if there are n objects, and if there is no restriction on the number of objects to be selected, we can choose the objects in 2^n ways. This includes 1 way of selecting 0 object. For exampl if you have 3 objects a, b, c, how many ways you can make the selection?:

No of ways to select 0 object - 1
Number of ways to select 1 object - 3 (a, b, c)
Number of ways to select 2 objects - 3 (ab, bc, ac)
Number of ways to select 3 objects -1 (abc)

Total = 8 = 2^3

In our example, we have H - 4, R - 5, AND C-3

Number of ways to select H = 2^4
Number of ways to select R = 2^5
Number of ways to select C = 2^3

But, we must select 1R and 1C, So we have to ignore 1 way of selecting 0 H and 1 way of selecting 0 C

So required number = (2^5 - 1) * (2^3 - 1) * (2^4) = 3472

Last edited by gmatblast on 05 Feb 2004, 12:15, edited 1 time in total.
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05 Feb 2004, 12:12
gmatblast:

Lucid Explanation. Thanks

I know nPr and nCr for selecting r objects out of n objects. But here how did you decide of applying 2^n formula for the selection of the objectds.

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05 Feb 2004, 12:17
pakoo wrote:
gmatblast:

Lucid Explanation. Thanks

I know nPr and nCr for selecting r objects out of n objects. But here how did you decide of applying 2^n formula for the selection of the objectds.

I think I explained this by way of an example. Please refer the bold text part in my posting. The bold text is just the explanantion of how 2^n is used. It has nothing to do with this problem. The solution of this problem starts only after the bold portion.
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05 Feb 2004, 22:08
Hey Gmatblast....ur expalanation of the topic is toooo cool man......
hey any reference for studying Probability and Permuations (Combinations). I am very week in these 2 areas.
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06 Feb 2004, 07:30
mdfrahim wrote:
Hey Gmatblast....ur expalanation of the topic is toooo cool man......
hey any reference for studying Probability and Permuations (Combinations). I am very week in these 2 areas.

http://www.gmatclub.com/members/courses ... lity.shtml

For more study material, I would suggest that you pick up high school math book that covers probability.
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06 Feb 2004, 15:50
Hey gmat_blast...that was awesome.... for the explanation.

Vivek.
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"Start By Doing What Is Necessary ,Then What Is Possible & Suddenly You Will Realise That You Are Doing The Impossible"

06 Feb 2004, 15:50
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