M40-52

Question

A book publisher has 4 horror stories, 3 romantic stories, and 2 comedy stories for his upcoming book. If the book must include at least one romantic story and at least one comedy story, and the order of the stories in the book does not matter, how many different books can he publish?

A. 6
B. 21
C. 96
D. 128
E. 336

A. 6
B. 21
C. 96
D. 128
E. 336

Bunuel · Answer

Official Solution:

A book publisher has 4 horror stories, 3 romantic stories, and 2 comedy stories for his upcoming book. If the book must include at least one romantic story and at least one comedy story, and the order of the stories in the book does not matter, how many different books can he publish?

A. 6
B. 21
C. 96
D. 128
E. 336

A. 6
B. 21
C. 96
D. 128
E. 336

Each of the 4 horror stories has two options - either to be included in the book or not. Hence, the total number of ways to represent the 4 horror stories in the book is  2^4 . Note that this count includes the scenario where none of the horror stories is included.
 
Similarly, each of the 3 romantic stories has two options - either to be included in the book or not. Hence, the total number of ways to represent the 3 romantic stories in the book is  2^3 . However, since this count includes the scenario where none of the romantic stories is included, and we need at least one romantic story in the book, we should subtract that one case from  2^3 . Thus, the total ways to represent the 3 romantic stories in the book, ensuring at least one is included, is  2^3 - 1 .
 
Next, each of the 2 comedy stories has two options - either to be included in the book or not. Hence, the total number of ways to represent the 2 comedy stories in the book is  2^2 . However, since this count also includes the scenario where none of the comedy stories is included, and we need at least one comedy story in the book, we should subtract that one case from  2^2 . Thus, the total ways to represent the 2 comedy stories in the book, ensuring at least one is included, is  2^2 - 1 .
 
Therefore, the total number of ways to represent these stories in the book is  2^4(2^3 - 1)(2^2 - 1) = 336 .

Answer: E

dshin117 · Answer

Thanks for posting yet another great question! I was wondering whether you could tell me the flaw in my logic:

I assumed one romantic and one comedy were already selected. Therefore, the remaining 7 books have the option of being included or not included --> 2 ^ 7. Where did I go wrong?

Bunuel · Answer

Using 2^7 is incorrect because it assumes you're only making combinations with one pre-selected romantic, say A, and one pre-selected comedy story, say X. This method generates only the combinations that include both A and X. However, we can have combinations that do not include A or X, or both. Therefore, this method reduces the total number of possible combinations. You need to use the method shown above to account for all possible story combinations.

einstein801 · Answer

What's wrong with doing 3C1*2C1*2^7?  (Choose 1 romantic out of 3, choose 1 comedy out of 2, remaining books have yes/no option = 2^7)

Bunuel · Answer

That number would have duplicates. If you select a romantic story, say A, with 3C1, and then select another romantic story, say B, with 2^7, you would have A, B, and some other stories. However, you would get the same set of stories if you select romantic story B with 3C1 and then story A with 2^7, along with the same remaining stories. Thus, you should understand and stick to the method shown in the official solution.

KrishnaVamsiP · Answer

I think this is a  high-quality  question and I agree with explanation.

M40-52

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