Question 151513



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{_{15}P_{9}=]{{{15!/(15-9)!}}} Plug in {{{n=15}}} and {{{r=9}}}




*[Tex \LARGE \textrm{_{15}P_{9}=]{{{15!/6!}}} Subtract {{{15-9}}} to get 6




Expand 15!
*[Tex \LARGE \textrm{_{15}P_{9}=]{{{(15*14*13*12*11*10*9*8*7*6*5*4*3*2*1)/6!}}}




Expand 6!
*[Tex \LARGE \textrm{_{15}P_{9}=]{{{(15*14*13*12*11*10*9*8*7*6*5*4*3*2*1)/(6*5*4*3*2*1)}}}




*[Tex \LARGE \textrm{_{15}P_{9}=]{{{(15*14*13*12*11*10*9*8*7*cross(6*5*4*3*2*1))/(cross(6*5*4*3*2*1))}}}  Cancel




*[Tex \LARGE \textrm{_{15}P_{9}=]{{{15*14*13*12*11*10*9*8*7}}}  Simplify





*[Tex \LARGE \textrm{_{15}P_{9}=]{{{1816214400}}}  Now multiply 15*14*13*12*11*10*9*8*7 to get 1,816,214,400



So 15 choose 9 (where order does matter) yields 1,816,214,400 unique combinations