swerve wrote: ↑

Tue Jan 19, 2021 4:50 pm

Roger wants to arrange three of his five books on his bookshelf. Two of the five books are duplicates and can not both be selected. In how many different ways can Roger arrange his books?

A. 12
B. 36
C. 42
D. 60
E. 128

The OA is C

Solution:

The number of ways to select 3 books from 5 books is 5C3 = (5 x 4 x 3)/3! = 10.

Out of these 10 ways, the number of ways that 2 of the 3 books will be duplicates is 2C2 x 3C1 = 1 x 3 = 3.

So there are 10 - 3 = 7 ways to select 3 books when the two duplicates are not selected.

For any 3 selected books, Roger can arrange them in 3! = 6 ways. Therefore, Roger can arrange the books in 7 x 6 = 42 ways.

Answer: C