muvjot wrote:

I would think this is a combination question because order does not matter. As long as 2 books are picked, it doesnt matter which book is picked. Is the answer given by the book wrong or someone please explain why?

QUESTION:

How many different ways can 4 books be arranged 2 at a time?

A. 15

B. 4

C. 24

D. 3

E. 12

ANSWER:

The order of arrangement matters in this problem. Therefore, this is a permutation problem. The number of permutations of n objects taken r at a time is:

P(n,r) = n!/(n-r)!

P(4,2) = 4!/(2!)

P(4,2) = 24/2

P(4,2) = 12

How many ways you can select 2 from 4C2 = 6

Each of this selection can be arrange in 2! ways = 2

Answer is 6*2 = 12

So it is a combintion and permutation problem.

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