Question 89825
Since order doesn't matter, we can use the combination formula:






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




*[Tex \LARGE \textrm{_{7}C_{3}=]{{{7!/(7-3)!3!}}} Plug in {{{n=7}}} and {{{r=3}}}




*[Tex \LARGE \textrm{_{7}C_{3}=]{{{7!/4!3!}}}  Subtract {{{7-3}}} to get 4



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



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




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




*[Tex \LARGE \textrm{_{7}C_{3}=]{{{(7*6*5)/3!}}}  Simplify



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




*[Tex \LARGE \textrm{_{7}C_{3}=]{{{210/(3*2*1)}}}  Multiply 7*6*5 to get 210




*[Tex \LARGE \textrm{_{7}C_{3}=]{{{210/6}}} Multiply 3*2*1 to get 6




*[Tex \LARGE \textrm{_{7}C_{3}=]{{{35}}} Now divide




So 7 choose 3 (where order doesn't matter) yields 35 unique combinations



So the answer is d)