Question 88733

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


*[Tex \LARGE \textrm{_{10}C_{8}=\frac{10!}{(10-8)!8!}}] Plug in {{{n=10}}} and {{{r=8}}}


*[Tex \LARGE \textrm{_{10}C_{8}=\frac{10!}{2!8!}}] Subtract {{{10-8}}} to get 2


*[Tex \LARGE \textrm{_{10}C_{8}=\frac{3628800}{2!8!}}] Calculate 10! to get 3,628,800 (note: if you need help with factorials, check out this <a href=http://www.algebra.com/algebra/homework/Probability-and-statistics/factorial.solver>solver</a>)


*[Tex \LARGE \textrm{_{10}C_{8}=\frac{3628800}{(2)(8!)}}] Calculate 2! to get 2


*[Tex \LARGE \textrm{_{10}C_{8}=\frac{3628800}{(2)(40320)}}] Calculate 8! to get 40,320



*[Tex \LARGE \textrm{_{10}C_{8}=\frac{3628800}{80640}}] Multiply the values 2 and 40,320 to get 80,640


*[Tex \LARGE \textrm{_{10}C_{8}=45] Divide 3,628,800 by 80,640 to get 45



So 10 choose 8 (where order does not matter) yields 45 unique combinations