JeffTargetTestPrep wrote:
AkshdeepS wrote:
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?
1. 12
2. 36
3. 42
4. 60
5.128
1. 12
2. 36
3. 42
4. 60
5.128
The number of ways to select 3 books when the two duplicates are selected is 2C2 x 3C1 = 1 x 3 = 3.
The number of ways to select 3 of 5 books is 5C3 = (5 x 4 x 3)/3! = 10.
So there are 10 - 3 = 7 ways to select 3 books when the two duplicates are not selected.
Since
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