EA student has 3 math books, 4 history books, 2 chemistry books, and 1 Latin boo

Question

EA student has 3 math books, 4 history books, 2 chemistry books, and 1 Latin book. He wants to arrange them on a bookshelf.

(Throughout B-E assume all books from a particular topic have the same title)

A. If all the books have different titles, in how many distinct ways can he arrange them?

B. In how many distinct ways can he arrange his books?

C. If he groups the identical math books together, and he groups the identical history books together, and he groups the identical chemistry books together, in how many distinct ways can he arrange his books?

D. If he groups the identical history books together (but isn't picky about the other books), in how many distinct ways can he arrange his books? E. What is the probability that none of the history books are together?

E. What is the probability that none of the history books are together?

Explanation / Answer

A. In total there are 10 books with different titles. So number of ways of arranging them = 10! = 3628800

B. Now there are r same titles, r=1,2,3,4 as books of same subject have same title.
So, number of ways of arranging books = 10! / (4!*3!*2!*1!) = 12600

C. The books of same subjects are arranged together. So, now number of distinct blocks of books = 4, on for each subject. So, number of ways of arranging = 4! = 24

D. 4 history books are grouped together as one block. So, number of blocks now = 1+2+3+1 = 7, out of which 2 chemistry books are similar and 3 maths books are similar. So, number of ways of arranging books = 7!/(2!*3!) = 420

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.