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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
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Bunuel
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

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
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