Question 1140100
{{{10C7 /5P3 }}}

first find {{{10C7}}}


{{{10C7=n!/((n-r)!r!)}}}

{{{10C7=10!/((10-7)!7!)}}}

{{{10C7=10!/(3!7!)}}}

{{{10C7=(10*cross(9)3*cross(8)4*cross(7*6*5*4*3*2*1))/(cross(3)*cross(2)*1*cross(7*6*5*4*3*2*1))}}}

{{{10C7=10*3*4}}}

{{{10C7=120}}}

then

{{{5P3=n!/(n-r)!}}}

{{{5P3=5!/(5-3)!}}}

{{{5P3=5!/2!}}}

{{{5P3=(5*4*3*cross(2*1))/(cross(2)*1)}}}

{{{5P3=5*4*3}}}

{{{5P3=60}}}



=> {{{10C7 /5P3 =120/60=2}}}