I need to understand how to do permutations.

Question

For the problem 3P2=(n!)/(n-r)! where n=3 and r=2, (The 3 and 2 at the beginning are subscript.) the book says the answer is 6 but I am getting 3. What am I doing wrong?

Answer 1

You might have using the formula for combinatoric instead of permutations. 

 

3C2 = 3!/((3-2)!2!) = 3 (wrong answer ...)

Answer 2

nPr=(n!)/(n-r)!

 

substitute values for n and r

 

3P2= 3!/(3-2)! = 3!/1! = 3*2*1/1 = 6