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Difficulty:
45%
(medium)
Question Stats:
68% (01:48) correct
32%
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wrong
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A book store has received 8 different books, of which 3/8 are novels, 25% are study guides and remaining are textbooks. If all books must be placed on one shelf displaying new items and if books in the same category have to be shelved next to each other, how many different arrangements of books are possible?
A. 18
B. 36
C. 72
D. 216
E. 432
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A. 18
B. 36
C. 72
D. 216
E. 432
_________________
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Answer: [E]
novels = 3
study guides = 2
textbooks = 3
total ways = 3! * 3! * 2! * 3! = 432 [arrangement between categories (3!) * individual arrangements ]
novels = 3
study guides = 2
textbooks = 3
total ways = 3! * 3! * 2! * 3! = 432 [arrangement between categories (3!) * individual arrangements ]
Posts: 13
There are 3 Novels, 2 Study Guides, and 3 Text Books.
The question specifies that books in the same category have to be shelved next to each other. This gives us 3! in addition to all of the different arrangements within each category.
Thus, Novel arrangements * Study Guide arrangements * Text Book arrangements * Category arrangements on the shelf = 3! * 2! * 3! * 3! = 432
The question specifies that books in the same category have to be shelved next to each other. This gives us 3! in addition to all of the different arrangements within each category.
Thus, Novel arrangements * Study Guide arrangements * Text Book arrangements * Category arrangements on the shelf = 3! * 2! * 3! * 3! = 432
Total Books : 8
Novels = 3/8 of 8 = 3
Study Guides = 25% = 1/4 of 8 = 2
Text books = 8 - 5 = 3
Arranging Novels , Study Guides and Text Books : NST ( 3! ways)
=> N itself has 3 different books which can be arranged in 3! ways
=> S has 2 different books which can be arranged in 2! ways
=> Text books can be arranged in 3! ways
so total arrangement possible : 3! * 3! * 2! * 3! = 6 * 6 * 2* 6 = 432
Novels = 3/8 of 8 = 3
Study Guides = 25% = 1/4 of 8 = 2
Text books = 8 - 5 = 3
Arranging Novels , Study Guides and Text Books : NST ( 3! ways)
=> N itself has 3 different books which can be arranged in 3! ways
=> S has 2 different books which can be arranged in 2! ways
=> Text books can be arranged in 3! ways
so total arrangement possible : 3! * 3! * 2! * 3! = 6 * 6 * 2* 6 = 432
JeffTargetTestPrep
Expert
Target Test Prep Representative
Joined: 04 Mar 2011
Last visit: 05 Jan 2024
Posts: 2,973
Status:Head GMAT Instructor
Affiliations: Target Test Prep
Expert reply
Posts: 2,973
Bunuel
The store has 3/8 x 8 = 3 novels, 1/4 x 8 = 2 study guides, and 3 textbooks.
We can let n = a novel, s = study guide, and t = textbook.
Since the each category of book has to be next to each other, we have:
[n(1)-n(2)-n(3)] [s(1)-s(2)] [t(1)-t(2)-t(3)]
We have 3 categories of books (novels, study guides, and textbooks), and so those categories of books can be arranged in 3! = 6 ways.
Now, let’s arrange the books within each category. The novels can be arranged in 3! = 6 ways, the study guides in 2! = 2 ways, and the textbooks in 3! = 6 ways.
Thus, the total number of ways to arrange the books is 6 x 6 x 2 x 6 = 432 ways.
Answer: E
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Using Word Permutation to solve this
Novels = 3/8*8 = 3
Study Guides = 8*25/100 = 2
Text books = 8 - 3 -2 = 3
Arranging the all books category wise = 3!
Arranging novel category books amongst themselves = 3!
Arranging text book category books amongst themselves = 3!
Arranging study guide category books amongst themselves = 2!
Total = 3! * 3! * 3! * 2! = 432
Novels = 3/8*8 = 3
Study Guides = 8*25/100 = 2
Text books = 8 - 3 -2 = 3
Arranging the all books category wise = 3!
Arranging novel category books amongst themselves = 3!
Arranging text book category books amongst themselves = 3!
Arranging study guide category books amongst themselves = 2!
Total = 3! * 3! * 3! * 2! = 432
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