Question 924801


{{{ C(n,r)=n!/((n-r)!2!) }}}


{{{ C(9,2)=9!/((9-2)!2!) }}}


{{{ C(9,2)=(9*8*7*6*5*4*3*2*1)/((7)!2! )}}}


{{{ C(9,2)=(9*8*7*6*5*4*3*2*1)/((7*6*5*4*3*2*1)(2*1))}}}


{{{ C(9,2)=(9*cross(8)4*cross(7*6*5*4*3*2*1))/(cross((7*6*5*4*3*2*1))(cross(2)*1))}}}


{{{ C(9,2)=9*4}}}


{{{ C(9,2)=36}}}


So {{{9}}} choose {{{2}}} (where order does not matter) yields {{{36}}} unique combinations.