Question 470766

Since order does not matter, we must use the <a href=http://www.mathwords.com/c/combination_formula.htm>combination formula</a>:



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




*[Tex \LARGE \textrm{_{8}C_{3}=]{{{8!/(8-3)!3!}}} Plug in {{{n=8}}} and {{{r=3}}}




*[Tex \LARGE \textrm{_{8}C_{3}=]{{{8!/5!3!}}}  Subtract {{{8-3}}} to get 5



Expand 8!
*[Tex \LARGE \textrm{_{8}C_{3}=]{{{(8*7*6*5*4*3*2*1)/5!3!}}}



Expand 5!
*[Tex \LARGE \textrm{_{8}C_{3}=]{{{(8*7*6*5*4*3*2*1)/(5*4*3*2*1)3!}}}




*[Tex \LARGE \textrm{_{8}C_{3}=]{{{(8*7*6*cross(5*4*3*2*1))/(cross(5*4*3*2*1))3!}}}  Cancel




*[Tex \LARGE \textrm{_{8}C_{3}=]{{{(8*7*6)/3!}}}  Simplify



Expand 3!
*[Tex \LARGE \textrm{_{8}C_{3}=]{{{(8*7*6)/(3*2*1)}}}




*[Tex \LARGE \textrm{_{8}C_{3}=]{{{336/(3*2*1)}}}  Multiply 8*7*6 to get 336




*[Tex \LARGE \textrm{_{8}C_{3}=]{{{336/6}}} Multiply 3*2*1 to get 6




*[Tex \LARGE \textrm{_{8}C_{3}=]{{{56}}} Now divide




So 8 choose 3 (where order doesn't matter) yields 56 unique combinations