Question 194347


Since order does matter, we must use the <a href=http://www.mathwords.com/p/permutation_formula.htm>permutation formula</a>:





*[Tex \LARGE \textrm{_{n}P_{r}=]{{{n!/(n-r)!}}} Start with the given formula




*[Tex \LARGE \textrm{_{13}P_{3}=]{{{13!/(13-3)!}}} Plug in {{{n=13}}} and {{{r=3}}}




*[Tex \LARGE \textrm{_{13}P_{3}=]{{{13!/10!}}} Subtract {{{13-3}}} to get 10




Expand 13!
*[Tex \LARGE \textrm{_{13}P_{3}=]{{{(13*12*11*10*9*8*7*6*5*4*3*2*1)/10!}}}




Expand 10!
*[Tex \LARGE \textrm{_{13}P_{3}=]{{{(13*12*11*10*9*8*7*6*5*4*3*2*1)/(10*9*8*7*6*5*4*3*2*1)}}}




*[Tex \LARGE \textrm{_{13}P_{3}=]{{{(13*12*11*cross(10*9*8*7*6*5*4*3*2*1))/(cross(10*9*8*7*6*5*4*3*2*1))}}}  Cancel




*[Tex \LARGE \textrm{_{13}P_{3}=]{{{13*12*11}}}  Simplify





*[Tex \LARGE \textrm{_{13}P_{3}=]{{{1716}}}  Now multiply 13*12*11 to get 1,716



So 13 choose 3 (where order does matter) yields 1,716 unique permutations



This means that there are 1,716 ways for 3 people to finish in the first three places.